🧠:[<think>Okay, let's try to tackle this problem. The task is to check that in 3-adic arithmetic, the number 201 is a square root of ...112101 with an accuracy of three digits. Hmm, so first, I need to recall how p-adic numbers work, especially 3-adic here. In p-adic arithmetic, numbers are represented in base p, which is 3 here, and they extend infinitely to the left. The accuracy of three digits probably means that when we square 201 (which is a 3-adic number), the last three digits should match the last three digits of ...112101. So let me confirm that.First, let's write down the number 201 in base 3. Wait, 201 is already in base 3? Wait, no, 201 in base 3 would mean 2*3² + 0*3 + 1, which is 2*9 + 0 + 1 = 19 in decimal. But in 3-adic numbers, the digits are written in base 3, so 201 is ...000201 in 3-adic, right? But since we are dealing with finite accuracy, maybe we just need to consider the last three digits.The number ...112101 in 3-adic. The last three digits are 101. So if we square 201 (in 3-adic), we need the last three digits of the square to be 101. Let's compute 201 squared in base 3.But how do we compute that? Let me convert 201 (base 3) to decimal first. Wait, maybe I should compute directly in base 3. Let's try that.First, 201 in base 3 is 2*3² + 0*3 + 1 = 2*9 + 0 +1 = 19 in decimal. If we square 19, we get 361. Now, convert 361 back to base 3. Let's see:3⁵=243, 3⁶=729. 361 is between 3⁵ and 3⁶.361 divided by 243 is 1 with a remainder of 118.118 divided by 81 (3⁴) is 1 with remainder 37.37 divided by 27 (3³) is 1 with remainder 10.10 divided by 9 (3²) is 1 with remainder 1.1 divided by 3 (3¹) is 0 with remainder 1.So 361 in base 3 is 1 1 1 1 0 1, which is 111101. But wait, the number we have is ...112101. Wait, but the problem says with an accuracy of three digits. The last three digits of ...112101 are 101, and 201 squared is 111101 in base 3, whose last three digits are 101. So that matches. Therefore, 201 squared in 3-adic arithmetic ends with 101, which is the last three digits of the target number. So that's the accuracy of three digits. Hence, 201 is a square root with that accuracy.But let me verify this computation step by step to make sure.First, let's compute 201 (base 3) squared directly in base 3. Multiplying numbers in base 3. Let's do it digit by digit.201*201------?In base 3:First, write 201 as each digit: 2 (3² place), 0 (3¹ place), 1 (3⁰ place).Multiplying 201 by 201 in base 3:Start with the rightmost digit of the multiplier (1):1 * 201 = 201Next digit is 0 (but it's the middle digit, so shifted one place):0 * 201 shifted one position: 0000Next digit is 2 (leftmost digit, shifted two places):2 * 201 shifted two positions: so 2*2=4 (in base 3, 4 is 11), but let's do the multiplication step by step.But maybe it's easier to compute the square step by step.Wait, maybe an easier way is to compute (2*3² + 1)^2.Expanding this: (2*3² + 1)^2 = (2*9 + 1)^2 = 19² = 361. But converting 361 to base 3 gives 111101 as before. Let's check:3⁵ = 243, 3⁶ = 729. 361 - 243 = 118 (1*3⁵)118 - 81 = 37 (1*3⁴)37 - 27 = 10 (1*3³)10 - 9 = 1 (1*3²)0 (0*3¹)1 (1*3⁰)So indeed, 111101 in base 3. The last three digits are 101. The given number is ...112101. The last three digits are 101, so they match. Therefore, 201 squared in 3-adic arithmetic ends with 101, which matches the last three digits of the given number. Therefore, 201 is a square root with an accuracy of three digits.But wait, the given number is ...112101. Let's see, if we have ...112101, then when we square 201, we get ...111101. The last three digits are 101, which matches. However, the fourth digit from the right is different: in ...112101, the fourth digit is 2, while in the square, it's 1. But since we're only checking accuracy of three digits, that's okay. The next digit (fourth) is not considered when accuracy is three digits. So, the three least significant digits (units, 3s, 9s places) must match, which they do (101).Therefore, the conclusion is correct. 201 is a square root of ...112101 in 3-adic arithmetic with an accuracy of three digits.Final Answerboxed{201}</think>To check that in 3-adic arithmetic, the number 201 is a square root of ...112101 with an accuracy of three digits, we need to verify that the last three digits of the square of 201 (in base 3) match the last three digits of ...112101.1. Convert 201 (base 3) to decimal: [ 201_3 = 2 cdot 3^2 + 0 cdot 3 + 1 = 2 cdot 9 + 0 + 1 = 19_{10} ]2. Square the decimal equivalent: [ 19_{10}^2 = 361_{10} ]3. Convert 361 (decimal) back to base 3: - (361 div 243 = 1) (remainder 118) - (118 div 81 = 1) (remainder 37) - (37 div 27 = 1) (remainder 10) - (10 div 9 = 1) (remainder 1) - (1 div 3 = 0) (remainder 1) - Therefore, (361_{10} = 111101_3)4. The last three digits of (111101_3) are 101. The given number ...112101 in 3-adic arithmetic also has the last three digits 101. Since the last three digits match, 201 is indeed a square root of ...112101 in 3-adic arithmetic with an accuracy of three digits.[boxed{201}],<think>Alright, let's tackle this problem. We need to check that in 3-adic arithmetic, the number 201 is a square root of ...112101 with an accuracy of three digits. Hmm, okay. Let me recall how p-adic numbers work. In 3-adic arithmetic, numbers are represented in base 3, and their expansions can go infinitely to the left. The "accuracy of three digits" probably means that when we square 201 (in 3-adic), the result should match ...112101 up to the first three digits from the right.First, let's convert 201 from base 3 to base 10 to see what we're dealing with. Wait, but in 3-adic numbers, 201 is actually a finite number, right? So 201 in base 3 is 2*3² + 0*3 + 1 = 2*9 + 0 + 1 = 19 in base 10. But the number ...112101 is a 3-adic number. Let's parse that. The rightmost digit is the 3⁰ place, then 3¹, 3², etc. So ...112101 would be, truncated to three digits, 101? Wait, but the problem says "with an accuracy of three digits," which I think means that when we square 201, the result should agree with ...112101 up to the first three 3-adic digits. So modulo 3³, which is 27. So we need to check that 201² ≡ ...112101 modulo 3³.Wait, but first, let's compute 201 squared in base 3. Let's do that step by step. 201 in base 3 is 2*9 + 0*3 + 1 = 19. So 19 squared is 361. Then, let's see what ...112101 is modulo 27 (since 3³ = 27). Let's convert ...112101 to base 10 modulo 27. The last three digits are 101 in base 3. Let's convert 101 (base 3) to base 10: 1*3² + 0*3 + 1 = 1*9 + 0 + 1 = 10. But the problem says the number is ...112101. Wait, so the last three digits are 101, which is 10 in base 10. But 201 squared is 19² = 361, and 361 modulo 27: 27*13 = 351, so 361 - 351 = 10. So 361 ≡ 10 mod 27. Which is exactly the same as the last three digits of ...112101 (which is 101 in base 3, equal to 10 in base 10). Therefore, 201 squared is congruent to ...112101 modulo 3³, which means it's accurate up to three digits in 3-adic arithmetic. Therefore, 201 is indeed a square root of ...112101 with accuracy of three digits.Wait, let me verify this again. Let me compute 201 (base 3) squared in base 3. Maybe converting to base 10 isn't necessary. Let's do the multiplication in base 3.201 * 201 in base 3:First, write 201 as: 2 0 1 * 2 0 1 ---------Multiply 1 (rightmost digit) by 201: 2 0 1 0 0 1 -------- 2 0 1Multiply 0 (middle digit) by 201, shifted one position: 2 0 1 0 0 0 -------- 0 0 0 0Multiply 2 (leftmost digit) by 201, shifted two positions: 2 0 1 2 0 1 0 0 -----------Now sum them up:First partial product: 201 (base 3)Second partial product: 0000 (base 3)Third partial product: 20100 (base 3)Adding these:Start from the right:First digit: 1Second digit: 0 + 0 = 0Third digit: 2 + 0 = 2Fourth digit: 0 + 1 = 1Fifth digit: 2So total is 2 1 2 0 1? Wait, that doesn't seem right. Let me perform the addition step by step.Adding 201 (base 3) and 20100 (base 3):Align the numbers: 0 2 0 1 + 2 0 1 0 0 ------------But actually, 201 is three digits, so when shifted, the partial products are:First partial product: 2 0 1 (three digits)Second partial product: 0 0 0 0 (four digits, shifted by one)Third partial product: 2 0 1 0 0 (five digits, shifted by two)Wait, perhaps the alignment is:First partial product (1 * 201): 2 0 1Second partial product (0 * 201 shifted by one): 0 0 0 0Third partial product (2 * 201 shifted by two): 2 0 1 0 0Now adding them:Start from the rightmost digit:1st digit: 12nd digit: 0 + 0 = 03rd digit: 2 + 0 = 24th digit: 0 + 1 = 1 (from the third partial product's 4th digit)5th digit: 2So combined, it's 2 1 2 0 1 (base 3). Wait, 21201 in base 3.Convert 21201 (base 3) to base 10:2*3^4 + 1*3³ + 2*3² + 0*3 + 1 = 2*81 + 1*27 + 2*9 + 0 +1 = 162 +27 +18 +1= 208.But earlier, 19 squared is 361. Wait, there's a discrepancy here. Wait, this suggests that multiplying 201 by 201 in base 3 gives 21201, which is 208 in base 10, but 19 squared is 361. That can't be. So something is wrong here.Wait, that must mean I made a mistake in the multiplication. Let's check again. 201 in base 3 is 2*9 + 0*3 +1 = 19. 19*19=361. So if the multiplication in base 3 is giving 21201 (which is 208), that's incorrect. Therefore, I messed up the multiplication steps.Let me perform the multiplication correctly. Let's multiply 201 (base 3) by 201 (base 3):First, rightmost digit is 1. Multiply 1 by 201: 201.Next, middle digit is 0. Multiply 0 by 201 and shift by 1: 0000.Next, leftmost digit is 2. Multiply 2 by 201 and shift by 2: 2*201 = 1120 (base 3). Wait, 2*2 in base 3 is 11 (since 2*2=4, which is 1*3 +1). Wait, actually, let's compute 2*201 in base 3:201* 2------Multiply each digit:1*2=2 (no carry)0*2=02*2=4. In base 3, 4 is 11. So write down 1 and carry over 1. But since there are no more digits, we write 11. So 2*201 in base 3 is 1100? Wait, no. Wait, 2*1 (rightmost digit) is 2. Then 2*0 is 0. Then 2*2 is 4. So, starting from the right:2 (units place), 0 (3s place), 4 (9s place). But 4 in 9s place is 1*3 +1, so we need to carry over. So 4 becomes 1*3 +1, so we write 1 and carry over 1. Then, the next digit (3s place) is 0 + carryover 1 =1. Then the 9s place is 1. So the result is 110 (base 3). Wait, but shifted by two positions because it's the leftmost digit. Wait, no. Wait, when multiplying the leftmost digit (2 in the 9s place), shifting by two positions would be equivalent to multiplying by 3². Wait, this is getting confusing. Maybe I should convert both numbers to base 10, multiply, and then convert back to base 3.201 (base 3) is 19 in base 10. 19*19=361. Convert 361 to base 3:3⁵=243, 3⁶=729. So 3⁵=243. 361 divided by 243 is 1 with remainder 118.118 divided by 81 (3⁴) is 1 with remainder 37.37 divided by 27 (3³) is 1 with remainder 10.10 divided by 9 (3²) is 1 with remainder 1.1 divided by 3 (3¹) is 0 with remainder 1.1 divided by 1 (3⁰) is 1.So reading the coefficients from highest to lowest: 1 (3⁵), 1 (3⁴), 1 (3³), 1 (3²), 0 (3¹), 1 (3⁰). Wait, but 3⁵=243, 3⁴=81, 3³=27, 3²=9, 3¹=3, 3⁰=1. So:243*1 +81*1 +27*1 +9*1 +3*0 +1*1 =243+81+27+9+0+1= 361. Correct.Therefore, 361 in base 3 is 111101. Wait, that's 1*3⁵ +1*3⁴ +1*3³ +1*3² +0*3¹ +1*3⁰. So 111101 (base 3). Therefore, 201 squared in base 3 is 111101 (base 3). But the problem states that ...112101 is the number we are taking the square root of. Wait, the example in the text later on says that 201 squared is 111101. The number ...112101 must have 111101 as part of its expansion? Wait, let's check.The given number is ...112101 in base 3. If we truncate it to six digits, it would be 112101. Let's convert 112101 (base 3) to base 10:1*3⁵ +1*3⁴ +2*3³ +1*3² +0*3 +1 = 243 +81 +54 +9 +0 +1= 243+81=324, 324+54=378, 378+9=387, 387+1=388. But 201 squared is 361, which is different. However, in the example, they mention that ...112101 minus 111101 (which is 201 squared) equals ...001000. So in 3-adic numbers, subtracting the finite part 111101 from ...112101 leaves ...001000, which is 3³ in 3-adic. Therefore, when they compute (201 + 2*10³)², they get ...112101 with higher accuracy.But back to the original problem. We need to check that 201 squared in 3-adic arithmetic is congruent to ...112101 modulo 3³. Wait, 3³ is 27, so we need to check that 201² ≡ ...112101 mod 27. But ...112101 mod 27 is equivalent to the last three digits 101 (base 3), which is 1*9 +0*3 +1=10. 201 squared in base 10 is 19²=361, which mod 27 is 361 - 13*27=361-351=10. So yes, 361 ≡10 mod27, which is 101 in base 3. But ...112101 mod27 is also 101 (base3)=10. Therefore, 201 squared ≡ ...112101 mod27. Hence, 201 is a square root of ...112101 with an accuracy of three digits.Therefore, the answer is correct. The key steps are converting both numbers modulo 3³ (27), checking that their squares match modulo 27, which they do (both equal 10 mod27). Therefore, 201 is indeed a square root with three-digit accuracy in 3-adic arithmetic.Final Answerboxed{201}</think>To check that in 3-adic arithmetic, the number 201 is a square root of ...112101 with an accuracy of three digits, we need to verify that squaring 201 (in 3-adic) results in a number congruent to ...112101 modulo (3^3) (27).1. Convert 201 (base 3) to base 10: [ 201_3 = 2 cdot 3^2 + 0 cdot 3 + 1 = 2 cdot 9 + 0 + 1 = 19_{10} ]2. Square 201 in base 10: [ 19^2 = 361_{10} ]3. Convert 361 (base 10) to base 3: [ 361_{10} = 1 cdot 3^5 + 1 cdot 3^4 + 1 cdot 3^3 + 1 cdot 3^2 + 0 cdot 3 + 1 = 111101_3 ]4. Check the last three digits of ...112101 (base 3): The last three digits of ...112101 are 101 (base 3), which converts to base 10 as: [ 101_3 = 1 cdot 3^2 + 0 cdot 3 + 1 = 9 + 0 + 1 = 10_{10} ]5. Verify congruence modulo 27: [ 361_{10} mod 27 = 361 - 13 cdot 27 = 361 - 351 = 10_{10} ] This matches the last three digits of ...112101 when converted to base 10 modulo 27.Thus, 201 squared in 3-adic arithmetic is congruent to ...112101 modulo (3^3), confirming that 201 is a square root of ...112101 with an accuracy of three digits.[boxed{201}]]

🔑:106. Indeed,begin{aligned}& begin{array}{r}201 times frac{201}{201}end{array} & 000 & 1102 & 111101end{aligned}The last three digits of the numbers 111101 and ... 112101 coincide.

🧠:[<think>Okay, so I need to find the volume bounded by the surfaces z = 2x² + 18y² and z = 6. Hmm, let's see. First, I remember that when calculating volumes between two surfaces, you can set up a double integral over the region where the two surfaces intersect. The volume would be the integral of the upper surface minus the lower surface over that region. Wait, in this case, z = 6 is a horizontal plane, and z = 2x² + 18y² is a paraboloid that opens upward. So, the region where these two surfaces intersect would be where 2x² + 18y² = 6. That means the paraboloid intersects the plane z = 6 along the curve 2x² + 18y² = 6. So, the volume we're looking for is under the plane z = 6 and above the paraboloid z = 2x² + 18y². To compute this volume, I need to set up a double integral over the region in the xy-plane where 2x² + 18y² ≤ 6. Then, the volume V would be the integral over this region of (6 - (2x² + 18y²)) dA. But integrating over this region directly might be a bit complicated because of the coefficients in front of x² and y². Maybe I can simplify the equation 2x² + 18y² = 6 to make it look like an ellipse equation. Let's divide both sides by 6: (x²)/3 + (y²)/(1/3) = 1. So, this is an ellipse with semi-axes lengths of sqrt(3) along the x-axis and sqrt(1/3) along the y-axis. Hmm, integrating over an elliptical region. I think changing variables to transform the ellipse into a circle would make the integral easier. That way, I can use polar coordinates. Let me recall how to do that. If I have an ellipse given by (x²)/(a²) + (y²)/(b²) = 1, a substitution that can be used is x = a*u and y = b*v, which transforms the ellipse into the unit circle u² + v² = 1. Then, the Jacobian determinant of this transformation is a*b, so dA = a*b du dv. Applying this to our case, where a² = 3 and b² = 1/3. Wait, so a is sqrt(3) and b is sqrt(1/3). Therefore, the substitution would be x = sqrt(3)*u and y = sqrt(1/3)*v. Then, the Jacobian determinant J is sqrt(3)*sqrt(1/3) = sqrt(1) = 1. Wait, that's nice! The Jacobian is 1, so the area element dA becomes du dv. That simplifies things. So, substituting x = sqrt(3) u and y = (1/sqrt(3)) v. Then, the region 2x² + 18y² ≤ 6 becomes 2*(3u²) + 18*( (1/3)v² ) ≤ 6. Let's compute that:2*3u² = 6u², and 18*(1/3)v² = 6v². So, 6u² + 6v² ≤ 6, which simplifies to u² + v² ≤ 1. Perfect, that's the unit circle. Now, let's express the integrand 6 - 2x² - 18y² in terms of u and v. First, 2x² = 2*(3u²) = 6u², and 18y² = 18*(1/3 v²) = 6v². Therefore, the integrand becomes 6 - 6u² - 6v² = 6(1 - u² - v²). So, the volume integral becomes the double integral over the unit circle u² + v² ≤ 1 of 6(1 - u² - v²) * 1 du dv, since the Jacobian is 1. Now, switching to polar coordinates for u and v would make sense here. Let's let u = r cosθ and v = r sinθ, so that u² + v² = r². The limits for r would be from 0 to 1, and θ from 0 to 2π. The area element in polar coordinates is r dr dθ. Therefore, the integral becomes:V = ∫₀²π ∫₀¹ 6(1 - r²) * r dr dθLet me compute that. First, factor out the constants:V = 6 ∫₀²π dθ ∫₀¹ (1 - r²) r drCompute the inner integral first: ∫₀¹ (1 - r²) r drLet me expand that: ∫₀¹ (r - r³) drIntegrate term by term:∫ r dr = (1/2)r² evaluated from 0 to 1: (1/2)(1) - (1/2)(0) = 1/2∫ r³ dr = (1/4)r⁴ evaluated from 0 to 1: 1/4 - 0 = 1/4Therefore, the inner integral is 1/2 - 1/4 = 1/4Then, multiply by 6 and the integral over θ:V = 6 * (1/4) * ∫₀²π dθ = (6/4) * [θ]₀²π = (3/2) * (2π - 0) = 3/2 * 2π = 3πWait, so the volume is 3π. Let me check the steps again to make sure I didn't make a mistake. First, the substitution: x = sqrt(3)u, y = (1/sqrt(3))v. Then, Jacobian determinant is indeed sqrt(3)*(1/sqrt(3)) = 1. So, dA = du dv. Then, transformed the integrand correctly. Then, converting to polar coordinates, got the integral 6(1 - r²) over the unit circle, which in polar coordinates becomes 6 ∫∫(1 - r²) r dr dθ. Then, split into 6 times (integral over θ) times (integral over r). The inner integral over r was correct: ∫₀¹ (r - r³) dr = [1/2 r² - 1/4 r⁴]₀¹ = 1/2 - 1/4 = 1/4. Then, 6 * 1/4 * 2π = 6*(1/4)*(2π) = 3π. Yes, that seems right. Alternatively, maybe I can check the volume of the paraboloid. Wait, the general formula for the volume under z = ax² + by² and up to z = h is (π h²)/(2 sqrt(ab)) ). Wait, is that a formula I remember? Let me think. Wait, for an elliptic paraboloid z = (x²/a²) + (y²/b²), the volume under z = h is (π a b h²)/2. Wait, let me verify that. If we have z = (x²/a²) + (y²/b²), then solving for the area at height z gives an ellipse with area π a b z. But integrating from z=0 to z=h would give the volume as ∫₀^h π a b z dz = (π a b h²)/2. But in our case, the equation is z = 2x² + 18y². Let's write this as z = 2x² + 18y² = (x²)/(1/√2)² + (y²)/(1/(3√2))². Wait, maybe not. Let's see. Alternatively, if we write z = 2x² + 18y² = 2(x² + 9y²) = 2(x² + (3y)^2). Let me define u = x and v = 3y. Then, z = 2(u² + v²). Then, the equation z = 6 becomes 2(u² + v²) = 6 → u² + v² = 3. Wait, so in terms of u and v, the volume would be similar to a paraboloid. But perhaps scaling variables again. Alternatively, let's see if our answer of 3π makes sense. If we have a paraboloid z = 2x² + 18y², and we're looking up to z = 6. The cross-section at z = 6 is an ellipse as we found earlier. Alternatively, using the formula for the volume under a paraboloid. If we can write the equation as z = k(x² + y²), then the volume up to height h is π h²/(2k). But in our case, the equation is z = 2x² + 18y². So, it's an elliptic paraboloid. So, the formula for the volume under z = ax² + by² up to z = h is (π h²)/(2√(ab)) ). Let's check that. So, if we have z = ax² + by². Let's perform a substitution similar to before. Let u = sqrt(a)x, v = sqrt(b)y. Then, z = u² + v². The Jacobian determinant is 1/(sqrt(a) sqrt(b)). The region z ≤ h becomes u² + v² ≤ h. The volume is then ∫∫ (h - (u² + v²)) * (1/(sqrt(a) sqrt(b))) du dv. Changing to polar coordinates: ∫₀^{2π} ∫₀^sqrt(h) (h - r²) * (1/(sqrt(ab))) r dr dθ. Compute the inner integral: ∫₀^sqrt(h) (h r - r³) dr = [ (h r²)/2 - r⁴/4 ] from 0 to sqrt(h) = (h*(h))/2 - (h²)/4 = (h²/2 - h²/4) = h²/4. Then, multiply by 2π and 1/sqrt(ab): (h²/4) * 2π / sqrt(ab) = (π h²)/(2 sqrt(ab)). So, the formula is V = π h² / (2 sqrt(ab)). In our case, z = 2x² + 18y², so a = 2, b = 18, h = 6. Plugging into the formula: V = π (6)^2 / (2 sqrt(2*18)) = 36π / (2 sqrt(36)) = 36π / (2*6) = 36π / 12 = 3π. So, that confirms our previous result. Therefore, the volume is indeed 3π. Therefore, the answer should be 3π, which in boxed form is boxed{3pi}.Final AnswerThe volume is boxed{3pi}.</think>To find the volume bounded by the surfaces ( z = 2x^2 + 18y^2 ) and ( z = 6 ), we first determine the region of intersection by solving ( 2x^2 + 18y^2 = 6 ). Dividing by 6, we get the ellipse equation ( frac{x^2}{3} + frac{y^2}{1/3} = 1 ).We use the substitution ( x = sqrt{3}u ) and ( y = frac{1}{sqrt{3}}v ), transforming the ellipse into the unit circle ( u^2 + v^2 = 1 ). The Jacobian determinant of this transformation is 1, so the area element remains ( du , dv ).The integrand ( 6 - (2x^2 + 18y^2) ) becomes ( 6 - 6u^2 - 6v^2 = 6(1 - u^2 - v^2) ). Converting to polar coordinates ( u = r costheta ) and ( v = r sintheta ), the integral becomes:[V = 6 int_0^{2pi} int_0^1 (1 - r^2) r , dr , dtheta]Evaluating the inner integral:[int_0^1 (r - r^3) , dr = left[ frac{1}{2}r^2 - frac{1}{4}r^4 right]_0^1 = frac{1}{2} - frac{1}{4} = frac{1}{4}]Then, the volume is:[V = 6 cdot frac{1}{4} cdot 2pi = 3pi]Thus, the volume bounded by the given surfaces is (boxed{3pi}).,<think>Okay, let's see. I need to find the volume bounded by the surfaces z = 2x² + 18y² and z = 6. Hmm, so these two surfaces intersect, and the volume is the region enclosed between them. I think the first step is to figure out the region over which these two surfaces intersect. That should give me the limits of integration for calculating the volume.The equation z = 2x² + 18y² is a paraboloid opening upwards, right? And z = 6 is a horizontal plane. So the intersection happens where 2x² + 18y² = 6. Let me solve this equation to find the boundary of the region in the xy-plane. If I divide both sides by 6, I get (x²)/3 + (y²)/(1/3) = 1. Wait, let me check that again.Starting with 2x² + 18y² = 6. Divide both sides by 6:(2x²)/6 + (18y²)/6 = 1 ⇒ (x²)/3 + (3y²)/1 = 1. So that's an ellipse in the xy-plane. The standard form of an ellipse is (x²/a²) + (y²/b²) = 1. Comparing, we get a² = 3 and b² = 1/3. Therefore, a = sqrt(3) and b = 1/sqrt(3). So the region of integration is an ellipse with semi-major axis sqrt(3) along the x-axis and semi-minor axis 1/sqrt(3) along the y-axis. To compute the volume between z = 6 and the paraboloid, I can set up a double integral over this elliptical region where the height at each point (x, y) is 6 - (2x² + 18y²). But integrating over an ellipse might be a bit tricky in Cartesian coordinates. Maybe it's easier to use a coordinate transformation to convert the ellipse into a circle, which would allow me to use polar coordinates. Let me think about how to do that.The equation of the ellipse is (x²)/3 + (y²)/(1/3) = 1. To transform this into a unit circle, I can use substitution variables. Let me set u = x / sqrt(3) and v = y / (1/sqrt(3)) = y*sqrt(3). Then the ellipse equation becomes u² + v² = 1, which is a unit circle in the uv-plane. This substitution is a linear transformation, so the Jacobian determinant will be constant. Let's compute the Jacobian matrix. The transformation is:u = x / sqrt(3)v = y * sqrt(3)So the partial derivatives are:du/dx = 1/sqrt(3), du/dy = 0dv/dx = 0, dv/dy = sqrt(3)Therefore, the Jacobian determinant is (1/sqrt(3))*(sqrt(3)) - 0*0 = 1. That's nice, the Jacobian determinant is 1, so the area element dx dy becomes du dv without any extra scaling. That simplifies things because when I change variables, I don't have to worry about the area scaling factor; it remains the same.So now, the integral becomes the integral over the unit circle in the uv-plane of [6 - 2x² - 18y²] times the Jacobian determinant (which is 1). But I need to express x and y in terms of u and v. Since u = x / sqrt(3), then x = u*sqrt(3). Similarly, v = y*sqrt(3), so y = v / sqrt(3).Let's substitute these into the integrand:6 - 2x² - 18y² = 6 - 2*(u*sqrt(3))² - 18*(v / sqrt(3))²= 6 - 2*(3u²) - 18*(v² / 3)= 6 - 6u² - 6v²= 6(1 - u² - v²)So the integrand simplifies to 6(1 - u² - v²). Therefore, the volume integral becomes:Volume = ∬_{u² + v² ≤ 1} 6(1 - u² - v²) du dvSince the region is a unit circle, switching to polar coordinates makes sense here. Let me convert to polar coordinates where u = r cosθ, v = r sinθ. The Jacobian determinant for polar coordinates is r, so du dv = r dr dθ. The limits of integration will be r from 0 to 1 and θ from 0 to 2π.Substituting into the integral:Volume = ∫₀^{2π} ∫₀¹ 6(1 - r²) * r dr dθFactor out the 6 and the θ integral:Volume = 6 * ∫₀^{2π} dθ * ∫₀¹ (1 - r²) r drFirst, compute the θ integral:∫₀^{2π} dθ = 2πThen compute the r integral:∫₀¹ (1 - r²) r dr = ∫₀¹ (r - r³) drIntegrate term by term:∫ r dr = (1/2)r² from 0 to 1 = 1/2∫ r³ dr = (1/4)r⁴ from 0 to 1 = 1/4So the r integral is (1/2 - 1/4) = 1/4Multiply everything together:Volume = 6 * 2π * (1/4) = 6 * (π/2) = 3πWait, hold on. Let me verify that again. So the θ integral is 2π, and the r integral is 1/4. Then 6 * 2π * 1/4 = (6 * 2 /4) * π = (12/4)*π = 3π. Hmm, so the volume is 3π?But let me check if I did everything correctly. Let me retrace the steps.1. Found the intersection curve by setting z=6 equal to the paraboloid. Got an ellipse.2. Performed substitution u = x/sqrt(3), v = y*sqrt(3). The Jacobian determinant was 1. So the area element is preserved.3. Expressed the integrand 6 - 2x² -18y² in terms of u and v, which became 6(1 - u² -v²).4. Converted to polar coordinates, leading to the integral 6∫∫(1 - r²) r dr dθ over the unit circle.5. Calculated the integrals step by step, resulting in 3π.Seems okay. But let me check the substitution step again. Because sometimes when you do variable substitution, even if the Jacobian is 1, you have to make sure the expressions are correctly transformed.Wait, let's check the integrand again. Starting with z=6 and z=2x² +18y². So the height is 6 - (2x² +18y²). Then substituting x = sqrt(3) u, y = v / sqrt(3). Then:2x² = 2*(3u²) = 6u²18y² = 18*(v² /3) = 6v²So 2x² +18y² =6u² +6v² =6(u² +v²)Therefore, 6 - (2x² +18y²)=6 -6(u² +v²)=6(1 -u² -v²). That's correct. So integrand is 6(1 - r²) in polar coordinates.Therefore, the integral is 6 times the integral over the unit circle of (1 - r²) r dr dθ. That is correct. The rest of the steps follow. The answer is 3π. Alternatively, maybe using another approach. For example, if I recall that the area of an ellipse is πab, where a and b are the semi-axes. In this case, the ellipse is (x²)/3 + y²/(1/3)=1, so a = sqrt(3) and b =1/sqrt(3). Then area is π*sqrt(3)*(1/sqrt(3))=π*(sqrt(3)/sqrt(3))=π. So area is π.But the volume here isn't just the area times height, because the height varies. The height is 6 - 2x² -18y². However, if we use the substitution and transform to the unit circle, we can also note that the integral over the unit circle of (1 - u² -v²) du dv can be calculated using polar coordinates as ∫∫(1 - r²) r dr dθ, which gives us (1/2 - 1/4) * 2π = π/2. Then multiplying by 6 gives 3π. So that's correct.Alternatively, if I didn't use substitution, could I have integrated in Cartesian coordinates? Let's see. The original integral would be over the ellipse. So the volume is ∬(6 -2x² -18y²) dx dy over the region 2x² +18y² ≤6.But integrating this in Cartesian coordinates would require limits for y dependent on x, which might be complicated. Let's attempt it.First, solve 2x² +18y² ≤6 for y:18y² ≤6 -2x² ⇒ y² ≤ (6 -2x²)/18 = (1/3) - (x²)/9So y is between -sqrt[(1/3 -x²/9)] and sqrt[(1/3 -x²/9)]Then x must satisfy 2x² ≤6 ⇒ x² ≤3 ⇒ x between -sqrt(3) and sqrt(3)Therefore, the integral becomes:∫_{-sqrt(3)}^{sqrt(3)} ∫_{-sqrt(1/3 -x²/9)}^{sqrt(1/3 -x²/9)} [6 -2x² -18y²] dy dxThis looks messy. Let me try to compute the inner integral with respect to y.Inner integral: ∫_{-sqrt(1/3 -x²/9)}^{sqrt(1/3 -x²/9)} [6 -2x² -18y²] dyFactor out constants:= (6 -2x²) * ∫_{-sqrt(1/3 -x²/9)}^{sqrt(1/3 -x²/9)} dy -18 ∫_{-sqrt(1/3 -x²/9)}^{sqrt(1/3 -x²/9)} y² dyFirst integral: (6 -2x²)* [2*sqrt(1/3 -x²/9)]Second integral: -18* [ (y³)/3 evaluated from -sqrt(1/3 -x²/9) to sqrt(1/3 -x²/9) ) ] = -18*[ ( (sqrt(1/3 -x²/9))³/3 - (-sqrt(1/3 -x²/9))³/3 ) ] = -18*[ (2*(1/3 -x²/9)^(3/2)/3 ) ] = -18*(2/3)*(1/3 -x²/9)^(3/2) = -12*(1/3 -x²/9)^(3/2)So the inner integral becomes:(6 -2x²)*2*sqrt(1/3 -x²/9) -12*(1/3 -x²/9)^(3/2)Simplify:Let me factor out (1/3 -x²/9)^(1/2) from both terms. Let’s denote A = sqrt(1/3 -x²/9). Then:First term: (6 -2x²)*2*ASecond term: -12*(A²)^(3/2) = -12*A³But A² =1/3 -x²/9, so A³ = A*(1/3 -x²/9)So:First term: 2*(6 -2x²)*ASecond term: -12*(1/3 -x²/9)*AFactor out 2*A:2*A*[ (6 -2x²) -6*(1/3 -x²/9) ]Compute the expression in brackets:First term: 6 -2x²Second term: -6*(1/3 -x²/9) = -2 + (2x²)/3So combining:6 -2x² -2 + (2x²)/3 = (6 -2) + (-2x² + (2x²)/3) = 4 - (4x²)/3Therefore, the inner integral is:2*A*(4 - (4x²)/3 )Substitute back A = sqrt(1/3 -x²/9):= 2*sqrt(1/3 -x²/9)*(4 - (4x²)/3 )Factor out 4:= 8*sqrt(1/3 -x²/9)*(1 - (x²)/3 )Therefore, the volume integral becomes:∫_{-sqrt(3)}^{sqrt(3)} 8*sqrt(1/3 -x²/9)*(1 - (x²)/3 ) dxThis still looks complicated. Let me try to simplify the expression under the integral.First, note that sqrt(1/3 -x²/9) can be written as sqrt( (3 -x²)/9 ) = (1/3)sqrt(3 -x²)Therefore:sqrt(1/3 -x²/9) = (1/3)sqrt(3 -x²)So substituting back into the integral:8*(1/3)sqrt(3 -x²)*(1 -x²/3 ) dx = (8/3)sqrt(3 -x²)*(1 -x²/3 ) dxLet’s compute the product inside:sqrt(3 -x²)*(1 -x²/3 ) = sqrt(3 -x²) - (x²/3)sqrt(3 -x²)So the integral becomes:(8/3) [ ∫_{-sqrt(3)}^{sqrt(3)} sqrt(3 -x²) dx - (1/3) ∫_{-sqrt(3)}^{sqrt(3)} x² sqrt(3 -x²) dx ]These integrals can be evaluated using standard techniques. Let's compute each part.First integral: ∫_{-a}^{a} sqrt(a² -x²) dx = (π a²)/2, which is the area of a semicircle. Here, a = sqrt(3), so the first integral is (π*(sqrt(3))²)/2 = (π*3)/2 = (3π)/2Second integral: ∫_{-sqrt(3)}^{sqrt(3)} x² sqrt(3 -x²) dxSince the integrand is even, we can compute from 0 to sqrt(3) and double it.Let’s let x = sqrt(3) sinθ, then dx = sqrt(3) cosθ dθ. When x=0, θ=0; x=sqrt(3), θ=π/2.Then:∫ x² sqrt(3 -x²) dx = ∫ (3 sin²θ) * sqrt(3 -3 sin²θ) * sqrt(3) cosθ dθ= ∫ 3 sin²θ * sqrt(3(1 - sin²θ)) * sqrt(3) cosθ dθ= ∫ 3 sin²θ * sqrt(3) cosθ * sqrt(3) cosθ dθ= ∫ 3 sin²θ * 3 cos²θ dθ= 9 ∫ sin²θ cos²θ dθUsing the identity sin²θ cos²θ = (1/4) sin²(2θ) = (1/8)(1 - cos4θ)So:9 * (1/8) ∫ (1 - cos4θ) dθ = (9/8)[ ∫ dθ - ∫ cos4θ dθ ]Integrate from 0 to π/2:= (9/8)[ θ - (1/4) sin4θ ] from 0 to π/2At π/2:θ = π/2; sin4θ = sin(2π) = 0At 0:θ =0; sin4θ=0Thus:= (9/8)[ (π/2 - 0) - (0 - 0) ] = (9/8)*(π/2) = (9π)/16But remember we had to double this because the original integral was from -sqrt(3) to sqrt(3)}. Wait, no. Wait, when we did substitution x = sqrt(3) sinθ, the limits changed from x=0 to x=sqrt(3) to θ=0 to θ=π/2, and we considered the integral from 0 to sqrt(3)}. Then we computed that integral as (9π)/16. Therefore, the original integral from -sqrt(3)} to sqrt(3)} is twice that? Wait no, because x^2 sqrt(3 -x²) is even function, so ∫_{-a}^a f(x) dx = 2 ∫_0^a f(x) dx. But in this case, we substituted x from 0 to sqrt(3)}, computed that integral as (9π)/16, so the total integral from -sqrt(3)} to sqrt(3)} is 2*(9π)/16 = 9π/8.Wait, wait, let me check again.Wait, original substitution:When x = sqrt(3) sinθ, dx = sqrt(3) cosθ dθ. So when x goes from 0 to sqrt(3)}, theta goes from 0 to π/2. The integral over x from 0 to sqrt(3)} becomes theta from 0 to π/2. Therefore, the integral ∫_{0}^{sqrt(3)} x² sqrt(3 -x²) dx = (9π)/16.But the integral from -sqrt(3)} to sqrt(3)} is twice that, because the integrand is even. So total is 2*(9π)/16 = 9π/8.Therefore, the second integral is 9π/8.Putting back into the expression:Volume = (8/3) [ (3π)/2 - (1/3)*(9π/8) ] = (8/3) [ (3π)/2 - (3π)/8 ]Compute inside the brackets:Convert to common denominator, which is 8:(3π)/2 = 12π/8, so 12π/8 - 3π/8 = 9π/8Thus, Volume = (8/3)*(9π/8) = (9π)/3 = 3πSo, same result. Therefore, confirms that the volume is indeed 3π.Alternatively, using another method: recognizing that the integral over the ellipse of (6 -2x² -18y²) dA can be evaluated by scaling coordinates.Given that the equation is 2x² +18y² =6. Let's factor out 6:2x² +18y² =6 ⇒ (x²)/(3) + (y²)/(1/3) =1. So as before.But another way is to perform a coordinate stretching to turn the ellipse into a circle.Suppose we let u = x / sqrt(3) and v = y / (1/sqrt(3)) as before. Then, as before, this transforms the ellipse into the unit circle. The area scaling factor is the Jacobian determinant, which we calculated as 1, but wait, actually, when I did the substitution earlier, u =x / sqrt(3), v = y*sqrt(3). So the Jacobian matrix is:du/dx =1/sqrt(3), du/dy=0dv/dx=0, dv/dy=sqrt(3)So Jacobian determinant is (1/sqrt(3))(sqrt(3)) =1, so the area element is preserved. Therefore, dx dy = du dv.Then, as before, the integrand becomes 6(1 -u² -v²), so integrating over the unit circle gives 6 times the volume under 1 -u² -v² over the unit circle. Which is 6*(the volume of a paraboloid subtracted from a cylinder). Wait, in this case, integrating (1 -u² -v²) over the unit circle is similar to finding the volume between z=1 -u² -v² and z=0, which is a paraboloid. The integral is the volume of that shape.But in any case, we computed that integral using polar coordinates and got 3π. And also verified via Cartesian coordinates. So seems solid.Therefore, the answer is 3π.Final AnswerThe volume is boxed{3pi}.</think>To calculate the volume of the body bounded by the surfaces ( z = 2x^2 + 18y^2 ) and ( z = 6 ), we first determine the region of intersection. Setting ( z = 6 ) equal to the paraboloid, we get:[2x^2 + 18y^2 = 6 implies frac{x^2}{3} + frac{y^2}{frac{1}{3}} = 1]This is an ellipse in the ( xy )-plane with semi-major axis ( sqrt{3} ) and semi-minor axis ( frac{1}{sqrt{3}} ).To simplify the integration, we use the coordinate transformation ( u = frac{x}{sqrt{3}} ) and ( v = ysqrt{3} ), which transforms the ellipse into a unit circle. The Jacobian determinant of this transformation is 1, so the area element ( dx , dy ) becomes ( du , dv ).The integrand ( 6 - 2x^2 - 18y^2 ) in terms of ( u ) and ( v ) is:[6 - 2(3u^2) - 18left(frac{v^2}{3}right) = 6 - 6u^2 - 6v^2 = 6(1 - u^2 - v^2)]Converting to polar coordinates ( u = r cos theta ) and ( v = r sin theta ), the integral becomes:[iint_{text{unit circle}} 6(1 - r^2) , du , dv = 6 int_0^{2pi} int_0^1 (1 - r^2) r , dr , dtheta]Evaluating the integrals:[int_0^{2pi} dtheta = 2pi][int_0^1 (1 - r^2) r , dr = int_0^1 (r - r^3) , dr = left[ frac{1}{2}r^2 - frac{1}{4}r^4 right]_0^1 = frac{1}{2} - frac{1}{4} = frac{1}{4}]Multiplying the results:[6 times 2pi times frac{1}{4} = 3pi]Thus, the volume of the body is (boxed{3pi}).]

🔑:## Solutionbegin{aligned}& a=frac{sqrt{z}}{sqrt{2}} ; b=frac{sqrt{z}}{sqrt{18}} & V=frac{1}{6} pi int_{0}^{6} z d z=left.frac{1}{12} pi z^{2}right|_{0} ^{6}=3 piend{aligned}Source — "http://pluspi.org/wiki/index.php/???????????? %D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD % mathrm{D} 1 % 82 % mathrm{D} 0 % mathrm{~B} % mathrm{D} 0 % mathrm{~B} 3 % mathrm{D} 1 % 80 % mathrm{D} 0 % mathrm{~B} 0 % mathrm{D} 0 % mathrm{BB} % mathrm{D} 1 % 8 mathrm{~B}+20-27 "Categories: Kuznetsov Problem Book Integrals Problem 20 | IntegralsUkrainian Banner Network- Last edited on this page: 16:14, 11 June 2009.- Content is available under CC-BY-SA 3.0.Created by GeeTeatoo## Problem Kuznetsov Integrals 20-28## Material from PlusPi

🧠:[<think>Alright, so I've got this problem here where I need to figure out something about the variables x, y, and z, which are defined in terms of a, b, c. The question is multiple choice, so I need to determine which of the statements A to D is correct. Let me start by understanding the problem step by step.First, let's write down the given equations:x = a² - 2b + π/3y = b² - 2c + π/6z = c² - 2a + π/2And the options are:A) At least one is greater than 0B) At least one is not less than 0C) At least one is not greater than 0D) At least one is less than 0Hmm. So we need to determine the truth values of these statements about x, y, z. Since a, b, c are real numbers, we need to analyze whether, regardless of the values of a, b, c, at least one of x, y, or z must satisfy each condition.Let me first parse the options:A) At least one of x, y, z is >0B) At least one is ≥0 (since "not less than 0" means ≥0)C) At least one is ≤0D) At least one is <0So, these are similar to different versions of the pigeonhole principle or perhaps using inequalities to bound these expressions. Maybe we can use some inequality manipulation here.Let me think. If I sum up all three equations, maybe I can find something about the sum x + y + z. If I can determine the sum's sign, that might help.Let me compute x + y + z:x + y + z = (a² - 2b + π/3) + (b² - 2c + π/6) + (c² - 2a + π/2)Combine like terms:= a² + b² + c² - 2b - 2c - 2a + (π/3 + π/6 + π/2)Simplify the constants:First, sum the π terms:π/3 + π/6 + π/2. Let's convert them to sixths:π/3 = 2π/6, π/6 = π/6, π/2 = 3π/6. So total is (2 + 1 + 3)π/6 = 6π/6 = π.So x + y + z = a² + b² + c² - 2a - 2b - 2c + πHmm. Let's see. Maybe complete the squares for each variable. For example:a² - 2a = (a - 1)^2 - 1Similarly, b² - 2b = (b - 1)^2 - 1c² - 2c = (c - 1)^2 - 1Therefore, substituting back into the sum:x + y + z = [(a - 1)^2 - 1] + [(b - 1)^2 - 1] + [(c - 1)^2 - 1] + πSimplify the constants:-1 -1 -1 + π = -3 + πAnd π is approximately 3.1416, so -3 + π ≈ 0.1416, which is positive.Therefore, x + y + z = (sum of squares) + (π - 3) ≈ sum of squares + 0.1416But each square term (a - 1)^2, (b - 1)^2, (c - 1)^2 is non-negative, so the entire sum x + y + z is equal to a non-negative number plus a positive number, which is positive. Therefore, x + y + z is positive.So, the sum of x, y, z is positive. Therefore, at least one of them must be positive, right? Because if all three were ≤0, their sum would be ≤0, which contradicts that the sum is positive. Therefore, that would mean that at least one of x, y, z is positive. So that would imply that statement A is true? Wait, but statement B is "at least one is not less than 0", which would include being equal to zero or positive. So if A is true, then B is also true. Hmm. But the options are exclusive? Wait, no, in multiple choice questions, sometimes more than one can be correct, but here the options are given as (A) to (D), and we need to pick the correct one among them. Wait, but in Chinese exams, often only one is correct. Let me check the original problem.Wait, the problem says "among x, y, z", and the options are labeled (A) to (D). The original problem probably expects exactly one correct answer. So need to see which one must be necessarily true.But according to the sum, since the sum is positive, at least one of them must be positive. So (A) is correct. But (B) is also correct because if something is greater than 0, it's also not less than 0. But maybe (B) is weaker. So if (A) is true, then (B) is automatically true. So perhaps the answer is (A). But wait, maybe (B) is also true. But the options are given as separate, so maybe the problem is designed to have only one correct answer. Hmm. Wait, but I need to check the problem statement again. The problem says "among x, y, z", then the options. So it's possible that multiple could be correct, but in the original problem, perhaps only one is. Let me check.Wait, the options are:(A) at least one is greater than 0(B) at least one is not less than 0(C) at least one is not greater than 0(D) at least one is less than 0Wait, but "not less than 0" is equivalent to "greater than or equal to 0". Similarly, "not greater than 0" is "less than or equal to 0". So, if (A) is true, then (B) is also true. However, in Chinese multiple choice questions, sometimes only the most precise answer is considered correct. So, if (A) is correct, then (B) is also correct but is a weaker statement, so (A) would be the better answer. But maybe not. Let me think again.Alternatively, maybe (A) is not necessarily true. Wait, but we have that the sum of x, y, z is positive, so at least one has to be positive. Therefore, (A) must be true. Then (B) is also true. But the problem might have multiple correct answers? Wait, the original problem is from an exam, probably with single answer. Let me check the problem again.Wait, let's think again. If the sum is positive, then at least one of them must be positive. So (A) is correct. But (B) says "at least one is not less than 0", which is equivalent to "at least one is ≥0". But if all of them are negative, then their sum would be negative, but the sum is positive, so indeed at least one must be positive. Therefore, (A) is true, which implies (B) is also true. But since (A) is a stronger statement, and the problem is structured as multiple choice with options A to D, perhaps both A and B are correct? But in the original problem statement, it's possible that the options are presented as separate, but in Chinese exams, typically only one is correct. Wait, maybe there's a mistake in my reasoning.Wait, let me check my calculation for the sum again. Let me redo the calculation step by step to be sure.Given:x = a² - 2b + π/3y = b² - 2c + π/6z = c² - 2a + π/2So sum x + y + z:= a² - 2b + π/3 + b² - 2c + π/6 + c² - 2a + π/2Combine like terms:a² + b² + c² - 2a - 2b - 2c + (π/3 + π/6 + π/2)Calculate the constants:π/3 + π/6 + π/2:Convert to sixths:π/3 = 2π/6π/6 = π/6π/2 = 3π/6Total: 2π/6 + π/6 + 3π/6 = 6π/6 = πSo sum is a² + b² + c² - 2a - 2b - 2c + πThen complete the squares:For each variable:a² - 2a = (a - 1)^2 -1b² - 2b = (b - 1)^2 -1c² - 2c = (c - 1)^2 -1Therefore, sum becomes:[(a - 1)^2 -1] + [(b - 1)^2 -1] + [(c - 1)^2 -1] + πCombine the constants:-1 -1 -1 + π = π - 3 ≈ 0.1416Thus, x + y + z = (a - 1)^2 + (b - 1)^2 + (c - 1)^2 + (π - 3)Since squares are non-negative, the entire sum is at least π - 3, which is approximately 0.1416, so positive.Therefore, x + y + z > 0. Therefore, at least one of x, y, z must be positive. Because if all were ≤0, the sum would be ≤0, which contradicts x + y + z >0. Therefore, at least one of x, y, z is positive. Therefore, option A is correct.But then option B says "at least one is not less than 0", i.e., at least one is ≥0. Since A is true (at least one is >0), then B is also true. However, in multiple choice questions, sometimes all correct answers must be selected, but here the options are (A)-(D), and in Chinese exams, typically only one is correct. So perhaps the answer is (A). Wait, but maybe I need to check if (A) is necessarily true. For example, is it possible that all three are exactly zero? But x + y + z ≈0.1416, which is positive, so they can't all be zero. So at least one must be positive. Therefore, (A) is correct.But wait, let's test some specific numbers. Let's choose a =1, b=1, c=1. Then compute x, y, z.x = 1² - 2*1 + π/3 = 1 - 2 + π/3 ≈ -1 + 1.047 ≈ 0.047 >0y = 1² - 2*1 + π/6 = 1 -2 + 0.523 ≈ -0.477 <0z = 1² -2*1 + π/2 = 1 -2 + 1.571 ≈ 0.571 >0So here x and z are positive, y is negative. So at least one is greater than 0, which is true.Another example: a=0, b=0, c=0.x = 0 -0 + π/3 ≈1.047 >0y = 0 -0 + π/6 ≈0.523 >0z=0 -0 + π/2 ≈1.571 >0All three positive. So A is true.Another case: Let me see if I can get two negative and one positive.Suppose a=2, b=2, c=2.x = 4 -4 + π/3 ≈0 +1.047≈1.047>0y=4 -4 + π/6≈0 +0.523≈0.523>0z=4 -4 +π/2≈0 +1.571≈1.571>0Still all positive. Hmm. Let me try different numbers.Suppose a=3, b=3, c=3.x=9 -6 +π/3≈3 +1.047≈4.047>0Similarly others would be even larger.Wait, maybe if I set a, b, c such that the linear terms dominate the squares. Let's see. For example, let a be very large negative. Suppose a=-10, then z = c² -2*(-10) + π/2 = c² +20 +1.571≈c² +21.571. That's still positive regardless of c. Similarly, x= (-10)^2 -2b + π/3≈100 -2b +1.047. Unless b is very large, x is positive. If b=60, then x=100 -120 +1.047≈-18.95, negative.So if I set a=-10, b=60, c=?Then compute x, y, z.x=100 -2*60 + π/3≈100 -120 +1.047≈-18.95 <0y= (60)^2 -2c + π/6≈3600 -2c +0.523≈3600.523 -2c. Unless c is extremely large, y is positive. Let's set c=2000.Then y=3600.523 -4000≈-399.47 <0z= (2000)^2 -2*(-10) + π/2≈4,000,000 +20 +1.571≈4,000,021.57>0So here x and y are negative, z is positive. So at least one is positive. So A holds.Alternatively, set a=1, b=1, c=1 as before, gives two positive, one negative. Another case: If a=1, b=1, c=1000.Then z = (1000)^2 -2*1 + π/2≈1,000,000 -2 +1.571≈999,999.571>0y=1 -2*1000 + π/6≈1 -2000 +0.523≈-1998.477<0x=1 -2*1 + π/3≈1 -2 +1.047≈0.047>0So again, x and z positive, y negative.Is there a case where only one is positive? Let me see. Let's try a=2, b=2, c=2.Wait, as before, all are positive. Let's think differently. Suppose we set a, b, c such that two variables are negative, but their squares might not compensate.Wait, but for example, let’s set a=0, b=0, c=0. Then all x, y, z are positive. If I set a=1, b=1, c=1, x≈0.047, y≈-0.477, z≈0.571. So x and z positive. If I set a=1, b=1, c=0.Then z =0 -2*1 + π/2≈-2 +1.571≈-0.429 <0y=1 -0 +π/6≈1 +0.523≈1.523>0x=1 -2*1 +π/3≈-1 +1.047≈0.047>0So here x and y positive, z negative. So again, two positive, one negative. If I set a=1, b=0, c=0.x=1 -0 +π/3≈1 +1.047≈2.047>0y=0 -0 +π/6≈0.523>0z=0 -2*1 +π/2≈-2 +1.571≈-0.429 <0Still two positive. Is there a way to get only one positive and two negative?Let me try a=3, b=3, c=3:x=9 -6 +1.047≈4.047>0y=9 -6 +0.523≈3.523>0z=9 -6 +1.571≈4.571>0All positive. If I set a=-1, b=-1, c=-1:x=1 -(-2) + π/3≈1 +2 +1.047≈4.047>0y=1 -(-2) +π/6≈1 +2 +0.523≈3.523>0z=1 -(-2) +π/2≈1 +2 +1.571≈4.571>0Still all positive. Hmm. Maybe it's difficult to get two negatives. Let me try a different approach.Wait, maybe even if two variables are negative, the squares would dominate. For example, let's set a=0.5, b=0.5, c=0.5.x=0.25 -1 +1.047≈0.25 -1 +1.047≈0.297>0y=0.25 -1 +0.523≈-0.227<0z=0.25 -1 +1.571≈0.821>0So two positives, one negative. If I set a=0.1, b=0.1, c=0.1:x=0.01 -0.2 +1.047≈0.857>0y=0.01 -0.2 +0.523≈0.333>0z=0.01 -0.2 +1.571≈1.381>0All positive. Hmm. So seems like unless variables are set to make the linear terms overpower the squares, but since squares grow faster, for large a, b, c, the squares will dominate. For example, if a is very large positive or negative, a² will dominate the linear term. Therefore, x, y, z will be positive for large |a|, |b|, |c|. The only way to make x, y, or z negative is when the linear term -2b, -2c, -2a overpowers the square and the constant term.For example, let's take x =a² -2b +π/3. If a is small, but b is large, then x could be negative. Similarly for y and z.But even if two variables are negative, the third might still be positive. So in all cases, at least one is positive. Therefore, A is correct.But then B is also correct because "not less than 0" includes being positive or zero. But if A is true, then B is also true, but since the problem is asking "among the options", and in the options, A and B are both there, but the answer should be A, because it's a stronger statement. However, in the problem statement, since sum is positive, it's possible that all three are positive, but even if not, at least one is positive. So A is necessarily true.But let's check whether B is also necessarily true. Well, if at least one is greater than 0, then that same one is also not less than 0. So B is a consequence of A. Therefore, both A and B are true. But since the problem is presented as a single-answer multiple choice, maybe the answer is both A and B, but in the options given, they are separate. Wait, looking back at the original problem:The options are:(A) at least one is greater than 0(B) at least one is not less than 0(C) at least one is not greater than 0(D) at least one is less than 0In Chinese exams, typically only one option is correct. So if A and B are both correct, but in the options, maybe only B is correct? Wait, but if A is correct, then B must be as well. But perhaps the answer is B. Wait, why?Wait, maybe there's a case where one of x, y, z is exactly zero. Then B would be true (since zero is not less than zero), but A would not necessarily be true. But according to our sum, x + y + z is equal to sum of squares plus π -3, which is approximately 0.1416. So if x + y + z is exactly π -3 ≈0.1416, then each square term must be zero. That is, a=1, b=1, c=1. Then x =1 -2 +π/3 ≈-1 +1.047≈0.047>0, y=1 -2 +π/6≈-1 +0.523≈-0.477<0, z=1 -2 +π/2≈-1 +1.571≈0.571>0. So even in that case, two are positive, one is negative. So there is no case where all are zero, and even if one of them is zero, the sum would require the others to compensate. But π -3 is approximately 0.1416, so even if two variables are negative, the third has to compensate more.But perhaps in some case, x=0, y=0, z=π-3≈0.1416>0. But to have x=0, a² -2b +π/3=0; y=0, b² -2c +π/6=0; z=c² -2a +π/2=π-3. Hmm, solving this system would require a lot of work. But even if x=0, since z>0, then at least one is positive, so A is still true.Wait, but suppose all x, y, z are zero. Then sum is zero, but we know sum is π -3≈0.1416≠0. So impossible. So it's impossible for all three to be zero or negative. Therefore, at least one must be positive, so A is correct, and since A is correct, B is also correct. However, in the problem, if only one answer is expected, then perhaps the answer is A.But maybe there is a trick here. Let's see. The problem says "among x, y, z". The options are about at least one being greater than 0, etc. But maybe there is a case where all are non-negative, making B correct but A not necessarily? But no, because if all are non-negative and their sum is positive, then at least one must be positive. So if all are non-negative, then B is true, but A would also be true. So if all are non-negative, both A and B are true. But if some are negative and some are positive, then A and B are still true. The only case where B is true but A is not is when at least one is zero, but since the sum is positive, others must compensate. For example, suppose x=0, y=0, z≈0.1416>0. Then A is still true because z>0, and B is also true. So in no case can B be true without A also being true. Therefore, A is a stronger statement, and since A must be true, B is redundant. But perhaps the answer is B? Wait, but B is "at least one is not less than 0", which is a weaker condition than A. So why would the answer be B?Wait, perhaps my initial assumption is wrong. Let me check again. Suppose all three are non-negative, then both A and B are true. If two are negative and one is positive, then A and B are true. If one is zero and others are positive or negative, then A and B are true. The only way B could be true without A is if one is exactly zero and others are negative. But if one is zero, then the sum x + y + z would be equal to the sum of the other two variables plus π -3. If two variables are negative and one is zero, then the sum would be negative + π -3. But π -3≈0.1416, so if the sum of the two negative variables is less than -0.1416, then the total sum would be negative, contradicting our previous conclusion that the sum is positive. Therefore, it's impossible for two variables to be negative and one zero. Because x + y + z must be positive. Therefore, if one variable is zero, the other two must sum to π -3 ≈0.1416. If one is zero, then the other two must sum to approximately 0.1416. But if one is zero, then the other two could be, for example, one positive and one negative, but their sum has to be positive. So even if one is zero, at least one of the other two must be positive. Therefore, in all cases, at least one variable is positive. Therefore, A must be true. Therefore, the answer is A.But why is option B present? Because in some cases, B is weaker. But since A implies B, but the problem is in Chinese and in Chinese exams, often the most accurate answer is required. So since A is always true, then A is the correct answer.But let me check again the original problem's options. The options are:(A) at least one is greater than 0(B) at least one is not less than 0(C) at least one is not greater than 0(D) at least one is less than 0But since the sum is positive, at least one is greater than 0. Therefore, A is correct.However, maybe there is a trick where even though the sum is positive, all three could be exactly zero? But we already saw that the sum is π -3≈0.1416, so they can't all be zero.Alternatively, maybe all three are non-negative but not all zero. Which would satisfy both A and B, but since the problem is a single choice, likely the answer is A.Alternatively, maybe the problem is designed to have answer B because even if one is zero, it's acceptable. But according to our previous analysis, even if one is zero, the sum requires others to be positive, so A still holds.Wait, but let's suppose that x=0, y=0, z=0.1416. Then, A is true because z>0, and B is also true because z≥0. But since x and y are zero, which are not less than zero, so B is also true. However, in this case, A is still true because z>0. Therefore, even in this case, A holds. Therefore, A is a better answer.But maybe there is a different line of reasoning. Suppose we assume that all three are less than or equal to zero. Then, x ≤0, y ≤0, z ≤0. Therefore, x + y + z ≤0. But we know that x + y + z = sum of squares + (π -3) >0. Therefore, a contradiction. Therefore, our assumption is wrong, so at least one of them must be greater than 0. Therefore, A is correct.Thus, the correct answer is A.But let me check the options again. The problem says:(A) at least one is greater than 0(B) at least one is not less than 0(C) at least one is not greater than 0(D) at least one is less than 0But in Chinese, sometimes the options might have nuances. But based on our analysis, A must be correct. But let's think again.Wait, the problem says "among x, y, z". Let's think about another approach: Suppose all three are less than or equal to zero. Then, x ≤0, y ≤0, z ≤0. Then, adding them up, x + y + z ≤0. But we know that x + y + z = sum of squares + π -3 ≈ sum of squares +0.1416. Since sum of squares is non-negative, x + y + z ≥0.1416>0. Contradiction. Therefore, at least one must be greater than 0. Therefore, A is correct.Therefore, the answer should be A.But looking at the options, why is B there? Because in some cases, perhaps someone might think that even if one is zero, it's okay. But since the sum is positive, at least one must be strictly positive. Therefore, A must be correct.Alternatively, perhaps the original problem's options are different. But given the current options, based on the sum being positive, we can conclude that at least one is greater than 0. Therefore, A is the answer.However, looking back at the problem statement, the equations for x, y, z include different constants: π/3, π/6, and π/2. These are approximately 1.047, 0.523, and 1.571. So if a², b², c² can be zero, then x would be -2b + π/3, y would be -2c + π/6, z would be -2a + π/2. So even if a, b, c are zero, x=π/3≈1.047>0, y=π/6≈0.523>0, z=π/2≈1.571>0. So in this case, all three are positive. If a, b, c are 1, as before, x≈0.047>0, y≈-0.477<0, z≈0.571>0. So depending on a, b, c, some can be negative. But the sum is always positive, so at least one is positive.Alternatively, maybe there's a way to have all x, y, z non-negative, but not necessarily all positive. But from the example where a=1, b=1, c=1, we saw that y is negative, but x and z are positive. So, in some cases, you can have negatives. But the key is that the sum is positive, so you can't have all non-positives.Wait, but the problem is about "among x, y, z", and the options are about the existence of at least one with some property. So according to our analysis, the answer is A. However, let me check the original problem's source. Wait, this seems like a problem similar to those in Chinese entrance exams, where often the correct answer is B. Wait, why?Wait, let's think differently. If we assume that the question is from an Olympiad, and the answer is B, then maybe the line of reasoning is different. Let me think. If all of x, y, z were negative, then their sum would be negative, but the sum is positive, so at least one must be non-negative (i.e., ≥0). Therefore, B is correct. But then if at least one is non-negative, it could be zero or positive. But the problem is asking "among the options", and both A and B are possible. But since the sum is positive, actually at least one must be positive. Therefore, A must be correct. But why would B be considered correct?Wait, maybe there's a translation issue. Let me check the original Chinese problem if possible. But since I don't have access to that, perhaps the answer is B. Wait, but according to the mathematical reasoning, since the sum is positive, at least one is positive. Therefore, A must be correct. Therefore, the answer should be A. However, perhaps the problem is in Chinese and the options are different. Wait, no, the user presented the problem in English. Then why is there confusion?Alternatively, maybe I made a mistake in calculating the sum. Let me check again:x = a² - 2b + π/3y = b² - 2c + π/6z = c² - 2a + π/2Sum:a² + b² + c² -2a -2b -2c + π/3 + π/6 + π/2Convert π terms:π/3 + π/6 = π/2, then π/2 + π/2 = π. So sum is a² + b² + c² -2a -2b -2c + π.Then complete the squares:(a-1)^2 -1 + (b-1)^2 -1 + (c-1)^2 -1 + πThus:(a-1)^2 + (b-1)^2 + (c-1)^2 + (π - 3)Since π - 3 ≈0.1416>0, so sum is positive. Therefore, the conclusion that at least one is positive is unavoidable. Therefore, A must be correct.Therefore, the answer is A. So I think the correct option is A.Final Answerboxed{A}</think>Given the problem where (a, b, c) are real numbers, and the expressions (x = a^2 - 2b + frac{pi}{3}), (y = b^2 - 2c + frac{pi}{6}), and (z = c^2 - 2a + frac{pi}{2}), we need to determine which of the statements A, B, C, or D is correct.First, we calculate the sum (x + y + z):[x + y + z = (a^2 - 2b + frac{pi}{3}) + (b^2 - 2c + frac{pi}{6}) + (c^2 - 2a + frac{pi}{2})]Combining like terms, we get:[x + y + z = a^2 + b^2 + c^2 - 2a - 2b - 2c + left(frac{pi}{3} + frac{pi}{6} + frac{pi}{2}right)]Simplifying the constants:[frac{pi}{3} + frac{pi}{6} + frac{pi}{2} = frac{2pi}{6} + frac{pi}{6} + frac{3pi}{6} = frac{6pi}{6} = pi]Thus, the sum becomes:[x + y + z = a^2 + b^2 + c^2 - 2a - 2b - 2c + pi]Next, we complete the square for each variable:[a^2 - 2a = (a - 1)^2 - 1][b^2 - 2b = (b - 1)^2 - 1][c^2 - 2c = (c - 1)^2 - 1]Substituting these into the sum:[x + y + z = [(a - 1)^2 - 1] + [(b - 1)^2 - 1] + [(c - 1)^2 - 1] + pi]Simplifying the constants:[x + y + z = (a - 1)^2 + (b - 1)^2 + (c - 1)^2 + (pi - 3)]Since (pi - 3 approx 0.1416) is positive, and each squared term is non-negative, the sum (x + y + z) is positive. Therefore, at least one of (x, y, z) must be positive. This implies that statement A ("at least one is greater than 0") is true.Given that the sum of (x, y, z) is positive, it is impossible for all three to be non-positive. Hence, at least one of them must be positive, making statement A correct. Thus, the final answer is:[boxed{A}],<think>Okay, let me see. So we have three real numbers a, b, c. Then we define x, y, z in terms of these a, b, c. The expressions for x, y, z are quadratic in a, b, c with some linear terms and constants. The question is about the signs of x, y, z. The options are whether at least one is greater than 0, at least one is not less than 0 (which means at least one is ≥0), at least one is not greater than 0 (≤0), or at least one is less than 0.Hmm. Let me first write down the equations to have them clear:x = a² - 2b + π/3y = b² - 2c + π/6z = c² - 2a + π/2So each variable is a quadratic function of one variable minus twice another variable plus a constant. The constants are π/3, π/6, and π/2. Let me note that π is approximately 3.1416, so π/3 is about 1.047, π/6 is about 0.523, and π/2 is about 1.571. So the constants are all positive, around 1.047, 0.523, and 1.571.So each of x, y, z is a combination of a square term (which is always non-negative), minus twice another variable, plus a positive constant. So perhaps we need to analyze the possible values of these expressions.The question is about the signs of x, y, z. The options are about whether at least one is positive (A), at least one is non-negative (B), at least one is non-positive (C), or at least one is negative (D). The options are phrased in terms of "at least one," so even if two of them are positive and one is negative, then (A) and (B) and (D) would all be true, but the answer options are exclusive. Wait, but actually, the options are separate, and the question is which of these statements is necessarily true among the four.Wait, but the problem is likely to have only one correct answer. Let me check the options again:(A) at least one is greater than 0(B) at least one is not less than 0 (so ≥0)(C) at least one is not greater than 0 (≤0)(D) at least one is less than 0But wait, (B) is a weaker statement than (A). Because if at least one is greater than 0, then that one is also not less than 0. So (A) implies (B). Similarly, (D) is stronger than (C). If at least one is less than 0, then that one is also not greater than 0, so (D) implies (C). So if (A) is true, then (B) is also true, but maybe the question is testing which of these must necessarily hold.But in the options, we have to select the correct one. Let's think.First, perhaps we can consider adding all three equations x + y + z. Maybe that's a way to start.x + y + z = (a² - 2b + π/3) + (b² - 2c + π/6) + (c² - 2a + π/2)Combine like terms:= a² + b² + c² - 2b - 2c - 2a + (π/3 + π/6 + π/2)Simplify the constants:π/3 + π/6 = π/2, so π/2 + π/2 = π. So total constant term is π.So x + y + z = a² + b² + c² - 2a - 2b - 2c + πHmm, this can be rewritten as:(a² - 2a) + (b² - 2b) + (c² - 2c) + πEach quadratic term can be completed to a square:a² - 2a = (a - 1)^2 - 1Similarly, b² - 2b = (b - 1)^2 - 1c² - 2c = (c - 1)^2 - 1Therefore:x + y + z = [(a - 1)^2 - 1] + [(b - 1)^2 - 1] + [(c - 1)^2 - 1] + πSimplify:= (a - 1)^2 + (b - 1)^2 + (c - 1)^2 - 3 + πSince π is approximately 3.1416, so π - 3 ≈ 0.1416, which is positive.Therefore, x + y + z = sum of squares (each non-negative) plus (π - 3). So sum of squares is non-negative, so x + y + z ≥ 0 + (π - 3) ≈ 0.1416 > 0Therefore, x + y + z is strictly positive.Therefore, the sum of x, y, z is positive. Therefore, at least one of x, y, z must be positive. Because if all three were non-positive, their sum would be non-positive, which contradicts that the sum is positive. Therefore, at least one of them must be positive. Therefore, option (A) is correct. Also, since (A) is correct, (B) is also correct because if at least one is greater than 0, then that one is also ≥0. But maybe the answer is (A). Wait, but let me check the options.Wait, the options are given as (A)-(D), and the question is which is necessarily true. So if (A) is correct, then (B) is also correct. But is that the case? Let me think. If all of x, y, z are non-negative, then (B) is true. If one is positive and others are negative, then (A) and (B) and (D) are true. If two are positive and one is negative, same. If all are positive, then (A) and (B) are true. However, the problem states that a, b, c are real numbers, so depending on the values of a, b, c, different scenarios may occur. But from the fact that x + y + z is positive, we can only conclude that at least one is positive. Therefore, (A) must be correct. But (B) is "at least one is not less than 0", which is equivalent to "at least one is ≥0". But if all x, y, z are non-negative, then (B) is true. But since we know that the sum is positive, so even if one is positive and the others are negative, (B) is true. But actually, if all x, y, z are positive, then (B) is true. So (B) is a weaker statement than (A). So if (A) is true, then (B) must also be true, but perhaps the answer is (A). Wait, but maybe the problem is a multiple choice where multiple options can be correct, but in the original problem statement, it's not specified. Wait, in the original problem, the options are (A)-(D), with the wording "among x, y, z" followed by the options. In Chinese exams, typically only one option is correct, but perhaps here as well. Wait, let me check again.Wait, the problem is given as a multiple-choice question with options (A)-(D). In such cases, usually only one option is correct, so we need to see which is necessarily true. However, the conclusion from the sum being positive is that at least one must be positive, so (A) is correct. But maybe there is a trick here.Alternatively, perhaps we can analyze each variable individually. Let's take x, y, z.Take x = a² - 2b + π/3. If we can show that x + y + z is positive, but each variable might not necessarily be positive? But how to know.Alternatively, let's suppose that all x, y, z are ≤0. Then x + y + z ≤0, but we just found that x + y + z >0, which is a contradiction. Therefore, at least one of x, y, z must be >0. Therefore, (A) is correct.Similarly, if we suppose that all x, y, z are <0, then their sum would be <0, which contradicts the sum being positive. Therefore, (D) is not necessarily true, but (A) is. Similarly, (B) is weaker than (A), so (B) is also true. Wait, but maybe in the problem statement, is it possible that (B) is the correct answer? Because if the problem says "at least one is not less than 0", which is equivalent to "at least one is ≥0", but since we have shown that at least one is >0, which implies that at least one is ≥0. So (A) implies (B). Therefore, if (A) is correct, (B) is also correct. But perhaps the answer is (A). However, in the options, (A) is "at least one is greater than 0" and (B) is "at least one is not less than 0". Since in mathematics, sometimes problems have multiple correct answers, but since the options are (A)-(D), and in Chinese exams, typically only one is correct, so perhaps (A) is the answer. Wait, but let's verify with some specific numbers.Suppose a = 1, b = 1, c = 1.Then x = 1 - 2*1 + π/3 ≈ 1 - 2 + 1.047 ≈ 0.047y = 1 - 2*1 + π/6 ≈ 1 - 2 + 0.523 ≈ -0.477z = 1 - 2*1 + π/2 ≈ 1 - 2 + 1.571 ≈ 0.571So x ≈ 0.047, y ≈ -0.477, z ≈ 0.571. Here, x and z are positive, y is negative.Therefore, in this case, (A) is true. Now, suppose we take a different case where maybe two are negative and one is positive.But from the sum being positive, even if two are negative, one must be sufficiently positive to make the sum positive.Alternatively, let's take a case where all three variables a, b, c are very large negative numbers. For example, let a = b = c = -1000.Then x = (-1000)^2 - 2*(-1000) + π/3 ≈ 1,000,000 + 2000 + 1.047 ≈ 1,002,001.047 >0Similarly, y = (-1000)^2 - 2*(-1000) + π/6 ≈ 1,000,000 + 2000 + 0.523 ≈ 1,002,000.523 >0z = (-1000)^2 - 2*(-1000) + π/2 ≈ 1,000,000 + 2000 + 1.571 ≈ 1,002,001.571 >0So all three are positive. So (A) is true.Alternatively, take a case where a, b, c are such that a² - 2b is minimized. Let's see, maybe set a = 1, b = 1, which makes x = 1 - 2 + π/3 ≈ 1 - 2 + 1.047 ≈ 0.047, which is positive. If we take a=0, b=0, c=0.Then x = 0 - 0 + π/3 ≈ 1.047 >0y = 0 - 0 + π/6 ≈ 0.523 >0z = 0 - 0 + π/2 ≈ 1.571 >0All positive again. So seems like in many cases, they are positive. But is there a case where x, y, z can all be negative? From the previous sum, since x + y + z is always positive, they cannot all be negative. So at least one must be positive. So (A) is correct.But the options are:(A) at least one is greater than 0(B) at least one is not less than 0 (so ≥0)(C) at least one is not greater than 0 (so ≤0)(D) at least one is less than 0From the sum, since it's positive, we can't have all ≤0, so (C) is not necessarily true. Similarly, we can't have all <0, so (D) is not necessarily true. But (A) must be true. (B) is also true, but since (A) is a stronger statement, if (A) is true, then (B) is redundant. But in the options given, (A) is the correct answer. But maybe in the problem, the answer is (A). Let's check if there's a possibility that (B) could be the correct answer. If all x, y, z are zero, then (B) is true, but in reality, since the sum is positive, at least one must be positive, so (A) must hold. Therefore, the answer should be (A). Wait, but looking back at the problem statement, maybe there's a trick here. Wait, the variables are a, b, c real numbers, so perhaps even if a, b, c are complex, but no, they are real. So the expressions x, y, z are real numbers.Alternatively, let's consider the possibility that one of x, y, z is zero. But even if one is zero, then (A) would still require another to be positive. But since the sum is positive, so even if two are zero, the third must be positive.But maybe another approach. Let's assume for contradiction that all x, y, z ≤0. Then:x = a² - 2b + π/3 ≤0y = b² - 2c + π/6 ≤0z = c² - 2a + π/2 ≤0So:a² - 2b ≤ -π/3b² - 2c ≤ -π/6c² - 2a ≤ -π/2Adding all three inequalities:a² + b² + c² - 2a - 2b - 2c ≤ - (π/3 + π/6 + π/2) = -πBut from before, when we computed x + y + z, we have:x + y + z = a² + b² + c² - 2a - 2b - 2c + πSo if we suppose x + y + z ≤0, then substituting:a² + b² + c² - 2a - 2b - 2c + π ≤0But from the inequalities above, if x, y, z ≤0, then a² + b² + c² - 2a - 2b - 2c ≤ -πSo substituting into x + y + z:(-π) + π ≤0 => 0 ≤0, which is equality. So the only case when x + y + z =0 is when the inequalities are equalities. But in reality, we have x + y + z = sum of squares + (π - 3) >0, since sum of squares is non-negative and π -3 ≈0.1416>0, so x + y + z >0 always. Therefore, assuming x, y, z ≤0 leads to a contradiction, so at least one must be >0. Therefore, (A) is correct.Thus, the answer is (A). But wait, looking back at the problem statement, the options are (A) at least one is greater than 0; (B) at least one is not less than 0; (C) at least one is not greater than 0; (D) at least one is less than 0.Since we have proved that at least one is greater than 0, (A) is correct. However, in the options, (B) is "at least one is not less than 0," which is equivalent to saying at least one is ≥0. But since we have proved that at least one is >0, then (A) implies (B). But depending on the problem's requirements, sometimes (B) is considered a weaker true statement, but in this case, since the sum is positive, (A) must hold, which is a stronger statement. Therefore, the correct answer is (A).Wait, but let me check once more with specific values where maybe one of x, y, z is exactly zero. Suppose x=0, then a² - 2b + π/3=0, so a² = 2b - π/3. Similarly, if possible. Let's try to set a=1, then 1 - 2b + π/3 =0 => 2b =1 + π/3 ≈1 +1.047≈2.047 => b≈1.0235. Then y = b² -2c + π/6 ≈(1.0235)^2 -2c +0.523 ≈1.047 -2c +0.523≈1.57 -2c. If we want y=0, then 1.57 -2c=0 => c≈0.785. Then z = c² -2a + π/2 ≈0.616 -2*1 +1.571≈0.616 -2 +1.571≈0.187. So in this case, x=0, y=0, z≈0.187>0. So at least one is positive, so (A) holds. If we try to set two variables to zero. Suppose x=0 and y=0. Then a² -2b +π/3=0 and b² -2c +π/6=0. Let me solve for a, b, c.From x=0: a²=2b -π/3From y=0: b²=2c -π/6 => c=(b² +π/6)/2Then z=c² -2a +π/2. Substitute c from above and a²=2b -π/3.So a=±√(2b -π/3). Let's pick a positive sqrt for simplicity. a=√(2b -π/3). Then z= [ (b² +π/6)/2 ]² - 2√(2b -π/3) + π/2.This is complicated, but let's plug in a value for b. Suppose we take b=1. Then a²=2*1 -π/3≈2 -1.047≈0.953, so a≈0.976. Then c=(1 + π/6)/2≈(1 +0.523)/2≈0.7615. Then z=(0.7615)^2 -2*0.976 +1.571≈0.579 -1.952 +1.571≈0.579 -1.952 is -1.373 +1.571≈0.198>0. So z is still positive. Therefore, even if we set x=0 and y=0, z is positive. Therefore, in all cases, at least one of x, y, z must be positive. Therefore, (A) is correct.Alternatively, if we tried to set z=0, then c² -2a +π/2=0 => c²=2a -π/2. Then a must be at least π/4≈0.785. Let's set a=1, then c²=2*1 -1.571≈0.429, so c≈0.655. Then x=1 -2b +1.047≈2.047 -2b. If we set x=0, then 2b=2.047 => b≈1.0235. Then y=(1.0235)^2 -2c +0.523≈1.047 -2*0.655 +0.523≈1.047 -1.31 +0.523≈0.26>0. So even if z=0, x=0, y is still positive. Therefore, in all cases, even if two variables are set to zero, the third is positive.Therefore, the conclusion is that at least one of x, y, z is greater than 0. So the correct answer is (A). Therefore, the answer is boxed{A}Final Answerboxed{A}</think>Given the problem where (a, b, c) are real numbers, and the expressions (x = a^2 - 2b + frac{pi}{3}), (y = b^2 - 2c + frac{pi}{6}), (z = c^2 - 2a + frac{pi}{2}), we need to determine which of the given options about the signs of (x, y, z) is correct.First, we consider the sum (x + y + z):[x + y + z = (a^2 - 2b + frac{pi}{3}) + (b^2 - 2c + frac{pi}{6}) + (c^2 - 2a + frac{pi}{2})]Combining like terms and simplifying the constants:[x + y + z = a^2 + b^2 + c^2 - 2a - 2b - 2c + left(frac{pi}{3} + frac{pi}{6} + frac{pi}{2}right)]The constants sum to (pi):[frac{pi}{3} + frac{pi}{6} + frac{pi}{2} = frac{pi}{2} + frac{pi}{2} = pi]Rewriting the quadratic terms by completing the square:[a^2 - 2a = (a - 1)^2 - 1, quad b^2 - 2b = (b - 1)^2 - 1, quad c^2 - 2c = (c - 1)^2 - 1]Thus, the sum becomes:[x + y + z = [(a - 1)^2 - 1] + [(b - 1)^2 - 1] + [(c - 1)^2 - 1] + pi]Simplifying further:[x + y + z = (a - 1)^2 + (b - 1)^2 + (c - 1)^2 - 3 + pi]Since (pi approx 3.1416), we have:[pi - 3 approx 0.1416 > 0]Therefore, the sum (x + y + z) is always positive. This implies that at least one of (x, y, z) must be positive. If all were non-positive, their sum would be non-positive, which contradicts the positivity of the sum. Hence, option (A) is correct.[boxed{A}]]

🔑:3. (A).begin{array}{l} x+y+z =left(a^{2}-2 b+frac{pi}{2}right)+left(b^{2}-2 c+frac{pi}{3}right)+left(c^{2}-2 aright. left.+frac{pi}{6}right) =left(a^{2}-2 a+1right)+left(b^{2}-2 b+1right)+left(c^{2}-2 c+1right) +(pi-3) =(a-1)^{2}+(b-1)^{2}+(c-1)^{2}+pi-3>0 . text { If } x leqslant 0, y leqslant 0, z leqslant 0, text { then } x+y+z leqslant 0 .end{array}begin{array}{l}text { This contradicts } x+y+z>0 . text { Therefore, at least one of } x, y, z text { is greater than } 0 .end{array}

🧠:[<think>Okay, let's see. I have a recursively defined sequence where the first two terms are 1 and 2, and then each subsequent term is the previous term divided by the one before that. So, a₁ = 1, a₂ = 2, and for n ≥ 1, a_{n+2} = a_{n+1} / a_n. The problem is asking for the sum of the first 1730 terms. Hmm, 1730 is a pretty large number. Maybe the sequence has a repeating pattern? That would make the sum easier to calculate if there's a cycle.Let me try to compute the first few terms manually to see if I can spot a pattern.Starting with a₁ = 1 and a₂ = 2.a₃ = a₂ / a₁ = 2 / 1 = 2a₄ = a₃ / a₂ = 2 / 2 = 1a₅ = a₄ / a₃ = 1 / 2 = 0.5a₆ = a₅ / a₄ = 0.5 / 1 = 0.5a₇ = a₆ / a₅ = 0.5 / 0.5 = 1a₈ = a₇ / a₆ = 1 / 0.5 = 2a₉ = a₈ / a₇ = 2 / 1 = 2a₁₀ = a₉ / a₈ = 2 / 2 = 1Wait a second, a₁₀ is 1, which is the same as a₁. And a₁₁ would be a₁₀ / a₉ = 1 / 2 = 0.5, which is a₃. Hmm, so starting from a₁ again, the sequence seems to repeat every 6 terms. Let me check this.Original terms:a₁ = 1a₂ = 2a₃ = 2a₄ = 1a₅ = 0.5a₆ = 0.5a₇ = 1a₈ = 2a₉ = 2a₁₀ = 1a₁₁ = 0.5a₁₂ = 0.5a₁₃ = 1So yes, from a₇ onwards, it's repeating 1, 2, 2, 1, 0.5, 0.5, which is the same as the first six terms starting from a₁. Therefore, the period here is 6. So the sequence cycles every 6 terms. Let me confirm:First six terms: 1, 2, 2, 1, 0.5, 0.5Next six terms: 1, 2, 2, 1, 0.5, 0.5And so on. So the cycle length is 6. Therefore, every block of 6 terms adds up to the same sum. Let's compute the sum of one cycle.Sum of one cycle (terms a₁ to a₆):1 + 2 + 2 + 1 + 0.5 + 0.5 = (1 + 2 + 2 + 1) + (0.5 + 0.5) = 6 + 1 = 7Wait, 1 + 2 is 3, plus 2 is 5, plus 1 is 6, then 0.5 + 0.5 is 1, so total sum 7. Correct.Then each subsequent cycle (terms a₇ to a₁₂, etc.) also sums to 7. So if the cycle length is 6, then the total number of cycles in 1730 terms would be 1730 divided by 6.Let me compute 1730 ÷ 6. 6*288 = 1728, so 1730 = 6*288 + 2. So there are 288 complete cycles and 2 leftover terms.Therefore, the total sum is (288 * 7) + sum of the first 2 terms of the cycle.Sum of first two terms: a₁ + a₂ = 1 + 2 = 3.Hence total sum = 288*7 + 3.Calculating 288*7:200*7 = 140080*7 = 5608*7 = 56So 1400 + 560 = 1960; 1960 + 56 = 2016Then 2016 + 3 = 2019.So the sum of the first 1730 terms is 2019.But let me double-check the cycle length. Maybe I made a mistake here. Let's list the terms again:1 (a₁)2 (a₂)2 (a₃) = 2/11 (a₄) = 2/20.5 (a₅) = 1/20.5 (a₆) = 0.5/11 (a₇) = 0.5/0.52 (a₈) = 1/0.52 (a₉) = 2/11 (a₁₀) = 2/20.5 (a₁₁) = 1/20.5 (a₁₂) = 0.5/11 (a₁₃) = 0.5/0.5Yes, so starting from a₁, the sequence is 1,2,2,1,0.5,0.5,1,2,2,1,0.5,0.5,1,...So the cycle is indeed 6 terms: [1,2,2,1,0.5,0.5], then repeats. So the sum per cycle is 7. Then 1730 divided by 6 gives 288 cycles with a remainder of 2. So 288*6=1728, then two more terms: a₁ and a₂ of the next cycle, which are 1 and 2.Therefore, sum is 288*7 + 1 + 2 = 2016 + 3 = 2019.Wait, but let me verify again the cycle. Let's check a₁₃ is 1, which is same as a₇ and a₁. So starting at term 1, 7,13,... the term is 1. So the cycle is 6 terms.Therefore, calculation seems correct.But to be thorough, let me check one more cycle:a₇ = 1a₈ = 2a₉ = 2a₁₀ = 1a₁₁ = 0.5a₁₂ = 0.5Then a₁₃ = 1, which is same as a₇. So yes, cycle is 6 terms. So sum of each cycle is 1+2+2+1+0.5+0.5=7.Therefore, total sum is (number of cycles)*7 + sum of remaining terms.Number of cycles: 1730 divided by 6. 6*288=1728. 1730-1728=2. So 288 cycles, 2 remaining terms.The remaining terms would be terms 1729 and 1730. Since each cycle is 6 terms, term 1729 is the first term of a new cycle, which is 1, and term 1730 is the second term, which is 2. So their sum is 1+2=3.Hence total sum is 288*7 + 3 = 2016 + 3 = 2019.Therefore, the answer is 2019.Final Answerboxed{2019}</think>The sequence is defined recursively with (a_1 = 1), (a_2 = 2), and (a_{n+2} = frac{a_{n+1}}{a_n}). We need to find the sum of the first 1730 terms.First, we compute the initial terms to identify a repeating pattern:[begin{align*}a_1 &= 1, a_2 &= 2, a_3 &= frac{a_2}{a_1} = 2, a_4 &= frac{a_3}{a_2} = 1, a_5 &= frac{a_4}{a_3} = 0.5, a_6 &= frac{a_5}{a_4} = 0.5, a_7 &= frac{a_6}{a_5} = 1, a_8 &= frac{a_7}{a_6} = 2, a_9 &= frac{a_8}{a_7} = 2, a_{10} &= frac{a_9}{a_8} = 1, a_{11} &= frac{a_{10}}{a_9} = 0.5, a_{12} &= frac{a_{11}}{a_{10}} = 0.5.end{align*}]The sequence repeats every 6 terms: (1, 2, 2, 1, 0.5, 0.5). The sum of one cycle (6 terms) is:[1 + 2 + 2 + 1 + 0.5 + 0.5 = 7.]To find the sum of the first 1730 terms, we determine the number of complete cycles and remaining terms:[1730 div 6 = 288 text{ cycles with a remainder of } 2.]The sum of 288 complete cycles is:[288 times 7 = 2016.]The remaining 2 terms are the first two terms of the next cycle, which are (1) and (2). Their sum is:[1 + 2 = 3.]Thus, the total sum of the first 1730 terms is:[2016 + 3 = 2019.][boxed{2019}],<think>Okay, let's see. I have this recursive sequence where the first two terms are 1 and 2, and then each subsequent term is the previous term divided by the one before that. So, a₁ = 1, a₂ = 2, and for n ≥ 1, a_{n+2} = a_{n+1} / a_n. The question is asking for the sum of the first 1730 terms. Hmm, that's a lot of terms. Maybe there's a pattern or cycle that repeats, so I don't have to calculate all 1730 terms individually.Let me start by writing out the first few terms to see if I can spot a pattern. We know a₁ = 1 and a₂ = 2. Let's compute a₃:a₃ = a₂ / a₁ = 2 / 1 = 2Then a₄ = a₃ / a₂ = 2 / 2 = 1a₅ = a₄ / a₃ = 1 / 2a₆ = a₅ / a₄ = (1/2) / 1 = 1/2a₇ = a₆ / a₅ = (1/2) / (1/2) = 1a₈ = a₇ / a₆ = 1 / (1/2) = 2a₉ = a₈ / a₇ = 2 / 1 = 2a₁₀ = a₉ / a₈ = 2 / 2 = 1Wait a minute, this looks familiar. Let me list them out:n: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,...aₙ:1, 2, 2, 1, 1/2, 1/2, 1, 2, 2, 1,...Hmm, starting from n=1, the terms go 1, 2, 2, 1, 1/2, 1/2, 1, 2, 2, 1,... So it seems like after the first two terms, the sequence cycles every 6 terms? Let's check:From a₁ to a₆: 1, 2, 2, 1, 1/2, 1/2Then a₇ is 1, which is the same as a₁. Then a₈ is 2, same as a₂. Then a₉ is 2, same as a₃, and so on. So starting from a₁, the cycle is 1, 2, 2, 1, 1/2, 1/2, and then repeats. Wait, but the cycle seems to be 6 terms? Let me confirm:Starting from a₁: 1, 2, 2, 1, 1/2, 1/2Then a₇ is 1, a₈ is 2, etc., so yes, the cycle is 6 terms. Let me write down the cycle terms:Cycle: [1, 2, 2, 1, 1/2, 1/2]Then it repeats. Let's check the next few terms:a₁₁ = a₁₀ / a₉ = 1 / 2 = 0.5 (which is 1/2), same as a₅a₁₂ = a₁₁ / a₁₀ = (1/2) / 1 = 1/2, same as a₆a₁₃ = a₁₂ / a₁₁ = (1/2) / (1/2) = 1, same as a₇Yep, definitely repeating every 6 terms. So the cycle length is 6. That means the sequence is periodic with period 6 starting from the first term. So the sum over each full cycle is the sum of these 6 terms. Let's calculate that.Sum of one cycle: 1 + 2 + 2 + 1 + 1/2 + 1/2Calculating step by step:1 + 2 = 33 + 2 = 55 + 1 = 66 + 1/2 = 6.56.5 + 1/2 = 7So each cycle of 6 terms adds up to 7. Therefore, if the sequence repeats every 6 terms, then the total sum for N terms would be (number of complete cycles) * 7 + sum of the remaining terms.Given that we need the sum of the first 1730 terms, let's divide 1730 by 6 to find out how many complete cycles there are and the number of remaining terms.1730 divided by 6. Let me compute that.6*288 = 1728, since 6*200=1200, 6*80=480, 6*8=48; 1200+480=1680, 1680+48=1728. So 288 cycles account for 1728 terms. Then 1730 - 1728 = 2 terms remaining. So there are 288 complete cycles and 2 extra terms.Therefore, the total sum is 288*7 + sum of the first 2 terms of the cycle.Wait, but the cycles start at term 1. So the remaining 2 terms would be the first two terms of the cycle, which are 1 and 2. So their sum is 1 + 2 = 3.Hence, total sum is 288*7 + 3.Calculating 288*7:280*7 = 19608*7 = 561960 + 56 = 2016Then 2016 + 3 = 2019.Therefore, the sum of the first 1730 terms is 2019.But wait, let me verify once again to be sure. Let me check the cycle again. The cycle is 6 terms: 1, 2, 2, 1, 1/2, 1/2. Sum is 7. Then each cycle is 6 terms, starting from term 1. So term 1 to term 6 is cycle 1, term 7 to term 12 is cycle 2, etc. So the total number of cycles in 1728 terms is 1728 / 6 = 288. Then the next two terms, term 1729 and term 1730, which would be the first two terms of the cycle: term 1729 is equivalent to term 1 (since 1728 is a multiple of 6, so term 1728 is the last term of a cycle, which is 1/2. Then term 1729 is term 1 of the next cycle, which is 1, and term 1730 is term 2, which is 2. So sum of these two terms is 1 + 2 = 3. Then total sum is 288*7 + 3 = 2016 + 3 = 2019. That seems correct.But let me double-check with a small example. Let's take N=6. Sum should be 7. Let's compute:1 + 2 + 2 + 1 + 1/2 + 1/2 = 1 + 2 + 2 + 1 + 1 = 7. Yes, that's correct.Then N=7. The sum would be 7 (from first 6 terms) + 1 (term 7) = 8. According to our formula, 1 full cycle (6 terms) gives 7, then 1 remaining term (which is term 7, the first term of the cycle) is 1. So total sum 7 + 1 = 8. Correct.Similarly, N=8: 7 + 1 + 2 = 10. But according to our formula, number of cycles: 8 /6 = 1 cycle (6 terms), remainder 2 terms. So 7 + 1 +2 =10. Correct.So the formula seems to hold.Therefore, the sum is 288*7 +3=2019.Final AnswerThe sum of the first 1730 terms is boxed{2019}.</think>The sequence is defined recursively with (a_1 = 1), (a_2 = 2), and (a_{n+2} = frac{a_{n+1}}{a_n}). We need to find the sum of the first 1730 terms.First, we compute the initial terms to identify a pattern:[begin{align*}a_1 &= 1, a_2 &= 2, a_3 &= frac{a_2}{a_1} = 2, a_4 &= frac{a_3}{a_2} = 1, a_5 &= frac{a_4}{a_3} = frac{1}{2}, a_6 &= frac{a_5}{a_4} = frac{1}{2}, a_7 &= frac{a_6}{a_5} = 1, a_8 &= frac{a_7}{a_6} = 2, a_9 &= frac{a_8}{a_7} = 2, a_{10} &= frac{a_9}{a_8} = 1.end{align*}]We observe that the sequence repeats every 6 terms starting from (a_1): (1, 2, 2, 1, frac{1}{2}, frac{1}{2}).The sum of one full cycle (6 terms) is:[1 + 2 + 2 + 1 + frac{1}{2} + frac{1}{2} = 7.]To find the sum of the first 1730 terms, we calculate the number of complete cycles and the remaining terms:[1730 div 6 = 288 text{ cycles with a remainder of } 2 text{ terms}.]The sum of 288 complete cycles is:[288 times 7 = 2016.]The remaining 2 terms are the first two terms of the cycle, (1) and (2), whose sum is:[1 + 2 = 3.]Thus, the total sum of the first 1730 terms is:[2016 + 3 = 2019.]The sum of the first 1730 terms is (boxed{2019}).]

🔑:Solution. Let's find the first few terms of the sequence: a_{3}=a_{2} / a_{1}= 2 / 1=2 ; a_{4}=a_{3} / a_{2}=2 / 2=1 ; a_{5}=a_{4} / a_{3}=0.5 ; a_{6}=a_{5} / a_{4}=0.5 / 1=0.5 ; a_{7}=a_{6} / a_{5}=0.5 / 0.5=1 and a_{8}=a_{7} / a_{6}=1 / 0.5=2. Notice that the sequence turned out to be periodic, its first six terms 1 ; 2 ; 2 ; 1 ; 0.5 ; 0.5 will repeat infinitely. Since the sequence is periodic, with a period of six, and 1728: 6=288 and a_{1}+a_{2}+ldots+a_{6}=7, then sum_{i=1}^{1730} a_{i}=sum_{i=1}^{1728} a_{i}+a_{1729}+a_{1730}=288 cdot 7+1+2=2019. Answer: 2019.

🧠:[<think>Okay, so I need to find the trajectory equation of the midpoint M of segment PQ, where point P is fixed at (10, 0) and point Q is moving on the circle x² + y² = 16. Hmm, let's start by recalling what a midpoint is. The midpoint M between two points P(x₁, y₁) and Q(x₂, y₂) has coordinates ((x₁ + x₂)/2, (y₁ + y₂)/2). That seems right.Since P is (10, 0), let me denote Q as (x, y) which is on the circle x² + y² = 16. Then the midpoint M would have coordinates ((10 + x)/2, (0 + y)/2). So, if I let the coordinates of M be (h, k), then h = (10 + x)/2 and k = y/2. Wait, that's correct? Let me check. Yes, midpoint formula averages the x-coordinates and the y-coordinates.Now, since Q is on the circle x² + y² = 16, we can express x and y in terms of h and k. Let's solve for x and y from the midpoint equations. From h = (10 + x)/2, multiply both sides by 2: 2h = 10 + x, so x = 2h - 10. Similarly, from k = y/2, multiply both sides by 2: y = 2k. So, substituting these back into the equation of the circle x² + y² = 16, we get:(2h - 10)² + (2k)² = 16.Let me write that out:(2h - 10)² + (2k)² = 16.Expanding that:First, (2h - 10)² = (2h)² - 2*2h*10 + 10² = 4h² - 40h + 100.Then, (2k)² = 4k².So adding them together:4h² - 40h + 100 + 4k² = 16.Let's simplify this equation. Subtract 16 from both sides:4h² - 40h + 100 + 4k² - 16 = 0.Simplify the constants: 100 - 16 = 84, so:4h² - 40h + 84 + 4k² = 0.I can factor out a 4 from all terms:4(h² - 10h + 21 + k²) = 0.Divide both sides by 4:h² - 10h + 21 + k² = 0.Hmm, maybe rearranging terms:h² + k² - 10h + 21 = 0.Now, to write the equation in standard form, perhaps complete the square for the h terms. Let's group h² -10h and complete the square.h² -10h = (h - 5)^2 - 25. So replacing that in the equation:(h - 5)^2 - 25 + k² + 21 = 0.Simplify constants: -25 +21 = -4.So, (h - 5)^2 + k² - 4 = 0.Add 4 to both sides:(h - 5)^2 + k² = 4.Therefore, the equation is (h - 5)^2 + k² = 4. But since h and k are the coordinates of the midpoint M, we can replace h with x and k with y to get the standard form in terms of x and y. So the trajectory equation of point M is (x - 5)^2 + y² = 4.Wait, let me verify this. The original circle has radius 4, centered at the origin. The midpoint M's locus is a circle with radius 2, centered at (5, 0). That seems plausible. Because when Q moves around the original circle, the midpoint between Q and (10,0) should trace a circle whose center is halfway between (10,0) and the center of the original circle, which is (0,0). The midpoint between (10,0) and (0,0) is (5,0), and the radius would be half of the original radius, which is 2. So the radius of the new circle is 2, centered at (5,0). Therefore, the equation (x -5)^2 + y² = 4 is correct.Let me check with specific points. Suppose Q is at (4,0), which is on the original circle. Then midpoint M is ((10 +4)/2, (0 +0)/2) = (7,0). Plugging into (x -5)^2 + y²: (7 -5)^2 +0 = 4, which is correct.Another point: Q is at (0,4). Midpoint M is ((10 +0)/2, (0 +4)/2) = (5, 2). Plugging into the equation: (5 -5)^2 + (2)^2 = 0 +4 =4, which works.Another point: Q is at (-4,0). Midpoint M is ((10 -4)/2, 0) = (3,0). Plugging into equation: (3 -5)^2 +0 = 4, which is 4. Correct.Seems like all these points satisfy the equation. So the trajectory is indeed a circle with center (5,0) and radius 2. Therefore, the equation is (x -5)^2 + y² =4.Final AnswerThe trajectory equation of the midpoint ( M ) is boxed{(x - 5)^2 + y^2 = 4}.</think>Given point ( P(10,0) ), and point ( Q ) is a moving point on the circle ( x^2 + y^2 = 16 ). We need to find the trajectory equation of the midpoint ( M ) of ( PQ ).1. The coordinates of the midpoint ( M ) of ( PQ ) are given by: [ h = frac{10 + x}{2}, quad k = frac{y}{2} ] where ( Q(x, y) ) is a point on the circle ( x^2 + y^2 = 16 ).2. Solving for ( x ) and ( y ) in terms of ( h ) and ( k ): [ x = 2h - 10, quad y = 2k ]3. Substituting ( x ) and ( y ) into the equation of the circle ( x^2 + y^2 = 16 ): [ (2h - 10)^2 + (2k)^2 = 16 ]4. Expanding and simplifying the equation: [ (2h - 10)^2 + (2k)^2 = 16 4h^2 - 40h + 100 + 4k^2 = 16 4h^2 - 40h + 84 + 4k^2 = 0 h^2 - 10h + 21 + k^2 = 0 ]5. Completing the square for the ( h )-terms: [ (h - 5)^2 - 25 + k^2 + 21 = 0 (h - 5)^2 + k^2 - 4 = 0 (h - 5)^2 + k^2 = 4 ]6. Replacing ( h ) and ( k ) with ( x ) and ( y ) respectively, the trajectory equation of the midpoint ( M ) is: [ boxed{(x - 5)^2 + y^2 = 4} ],<think>Okay, so I have this problem here where there's a point P at (10, 0), and point Q is moving along the circle defined by the equation x² + y² = 16. I need to find the trajectory equation of the midpoint M of the segment PQ as Q moves around the circle. Hmm, let me try to figure this out step by step.First, let me recall what a midpoint is. The midpoint M of two points P(x₁, y₁) and Q(x₂, y₂) has coordinates ((x₁ + x₂)/2, (y₁ + y₂)/2). Right? So in this case, P is fixed at (10, 0), and Q is moving on the circle x² + y² = 16. So, Q has coordinates (x, y) such that x² + y² = 16.Let me denote the coordinates of Q as (x_q, y_q). Then, the coordinates of M, the midpoint, would be ((10 + x_q)/2, (0 + y_q)/2). So, M's coordinates are ((10 + x_q)/2, y_q/2).Now, since Q is on the circle x² + y² = 16, we have x_q² + y_q² = 16. But I need to express the coordinates of M in terms of their own variables and eliminate x_q and y_q. Let me denote the coordinates of M as (h, k). Then:h = (10 + x_q)/2 k = y_q/2From these equations, I can solve for x_q and y_q in terms of h and k:x_q = 2h - 10 y_q = 2kSince Q lies on the circle x² + y² = 16, substituting x_q and y_q into this equation should satisfy it. Let's do that:x_q² + y_q² = 16 (2h - 10)² + (2k)² = 16Expanding this equation:(4h² - 40h + 100) + 4k² = 16 4h² - 40h + 100 + 4k² = 16Let's subtract 16 from both sides to set the equation to zero:4h² - 40h + 100 + 4k² - 16 = 0 4h² - 40h + 84 + 4k² = 0Hmm, maybe simplify this equation by dividing all terms by 4 to make it simpler:h² - 10h + 21 + k² = 0So, h² + k² - 10h + 21 = 0Now, this is the equation in terms of h and k, which are the coordinates of the midpoint M. But usually, we write equations in terms of x and y. So, replacing h with x and k with y, the equation becomes:x² + y² - 10x + 21 = 0Alternatively, maybe rearrange terms to write it in the standard form of a circle. Let's complete the square for the x terms:x² - 10x + y² + 21 = 0Take the x terms: x² - 10x. To complete the square, take half of -10, which is -5, square it to get 25, then add and subtract 25:(x² - 10x + 25) - 25 + y² + 21 = 0 (x - 5)² + y² - 4 = 0 (x - 5)² + y² = 4So, the equation is a circle with center at (5, 0) and radius 2. Let me check if this makes sense.Original circle Q is on has radius 4 (since x² + y² = 16) centered at (0,0). Point P is at (10,0). The midpoints between P and Q should form a circle. The center of this new circle should be halfway between the center of the original circle (0,0) and point P (10,0). The midpoint between (0,0) and (10,0) is indeed (5,0). The radius should be half of the original circle's radius, because the midpoint moves half the distance that Q moves. Original radius is 4, so half is 2. So the new circle should have radius 2. That checks out with the equation we found: (x - 5)² + y² = 4. So that seems correct.Let me just verify with an example. Suppose Q is at (4, 0), which is on the original circle (since 4² + 0² = 16). Then the midpoint M would be ((10 + 4)/2, (0 + 0)/2) = (7, 0). According to the equation (x - 5)² + y² = 4, plugging in (7,0) gives (7 -5)^2 + 0 = 4, which is 4. Correct.Another point: Q at (0,4). Midpoint M is ((10 + 0)/2, (0 + 4)/2) = (5, 2). Plugging into the equation: (5 -5)^2 + (2)^2 = 0 + 4 = 4. Correct.Q at (-4, 0). Midpoint M is ((10 -4)/2, 0) = (3,0). Plug into equation: (3 -5)^2 + 0 = 4 + 0 = 4. Correct.So, all these test points work. Therefore, the trajectory equation of the midpoint M is (x -5)^2 + y^2 =4, or written out as x² + y² -10x +21 =0.Therefore, the answer should be boxed{(x - 5)^2 + y^2 = 4}Final AnswerThe trajectory equation of the midpoint ( M ) is boxed{(x - 5)^2 + y^2 = 4}.</think>Given point ( P(10,0) ), and point ( Q ) is a moving point on the circle ( x^2 + y^2 = 16 ). We need to find the trajectory equation of the midpoint ( M ) of ( PQ ) as ( Q ) moves on the circle.1. The coordinates of the midpoint ( M ) of ( PQ ) are given by: [ h = frac{10 + x_q}{2}, quad k = frac{y_q}{2} ] where ( Q ) has coordinates ( (x_q, y_q) ).2. Solving for ( x_q ) and ( y_q ) in terms of ( h ) and ( k ): [ x_q = 2h - 10, quad y_q = 2k ]3. Since ( Q ) lies on the circle ( x^2 + y^2 = 16 ), substituting ( x_q ) and ( y_q ) into this equation: [ (2h - 10)^2 + (2k)^2 = 16 ]4. Expanding and simplifying the equation: [ 4h^2 - 40h + 100 + 4k^2 = 16 ] [ 4h^2 + 4k^2 - 40h + 84 = 0 ] Dividing by 4: [ h^2 + k^2 - 10h + 21 = 0 ]5. Rewriting in terms of ( x ) and ( y ): [ x^2 + y^2 - 10x + 21 = 0 ]6. Completing the square for the ( x )-terms: [ (x - 5)^2 + y^2 - 4 = 0 ] [ (x - 5)^2 + y^2 = 4 ]The trajectory equation of the midpoint ( M ) is (boxed{(x - 5)^2 + y^2 = 4}).]

🔑:This problem involves finding the trajectory equation. To solve this, we can use the substitution method. Let M(x,y) and Q(a,b). Since M is the midpoint of PQ and P is (10,0), we can use the midpoint formula to get a=2x-10 and b=2y. Also, Q(a,b) is a moving point on the circle x^{2}+y^{2}=16. By substituting a=2x-10 and b=2y into x^{2}+y^{2}=16, we can find the equation that the coordinates of point M(x,y) satisfy. Simplifying this equation gives us the trajectory equation of point M.Solution steps:1. Let M(x,y) and Q(a,b).2. Since M is the midpoint of PQ and P is (10,0), use the midpoint formula to find a and b: a=2x-10 and b=2y.3. Substitute a=2x-10 and b=2y into the circle equation x^{2}+y^{2}=16: (2x-10)^{2}+(2y)^{2}=16.4. Simplify the equation: (x-5)^{2}+y^{2}=4.Therefore, the trajectory equation of the midpoint M of PQ is boxed{(x-5)^{2}+y^{2}=4}.

🧠:[<think>Okay, so I have this problem here where I need to convert the vector equation of a line into its coordinate form. The vector equation is given as (vec{r} = tvec{e}), where (vec{e}) is a unit vector that makes angles (alpha), (beta), and (gamma) with the coordinate axes Ox, Oy, Oz respectively. The goal is to show that the coordinate form of this line's equation is (frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}). Hmm, let's see how to approach this.First, let me recall that in three-dimensional space, a line can be represented parametrically using a point and a direction vector. Since the line passes through the origin, the point is (0, 0, 0), and the direction vector is (vec{e}). The unit vector (vec{e}) has components equal to the cosines of the angles it makes with each axis. So, if (alpha), (beta), and (gamma) are the angles with the x, y, and z-axes respectively, then the direction cosines are (cos alpha), (cos beta), and (cos gamma). Therefore, the unit vector (vec{e}) can be written as (vec{e} = (cos alpha, cos beta, cos gamma)). Now, the vector equation (vec{r} = tvec{e}) can be written in terms of coordinates. Let me denote (vec{r}) as the position vector (x, y, z). Then, substituting into the equation, we get:[(x, y, z) = t(cos alpha, cos beta, cos gamma)]This gives us the parametric equations for each coordinate:[x = t cos alpha y = t cos beta z = t cos gamma]So, each coordinate (x, y, z) is a scalar multiple of the corresponding direction cosine, with the scalar being the parameter t. Now, to eliminate the parameter t and write the equation in symmetric form (coordinate form), we can solve each equation for t and set them equal to each other. Let's try that.From the first equation: (t = frac{x}{cos alpha})From the second equation: (t = frac{y}{cos beta})From the third equation: (t = frac{z}{cos gamma})Since all three expressions equal t, they must be equal to each other. Therefore:[frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}]Which is exactly the coordinate form given in the problem. Wait, so that's straightforward? Let me check if there's a possibility of error here.Wait, the direction vector is a unit vector, so the direction cosines should satisfy (cos^2 alpha + cos^2 beta + cos^2 gamma = 1). But maybe that's not directly relevant here. The key is that since (vec{e}) is a unit vector, the components are indeed the cosines of the angles, so this parametric form is valid.Alternatively, if the direction vector wasn't a unit vector, the components would be proportional to the direction cosines but scaled by the magnitude. But here, since it's a unit vector, the components are exactly the direction cosines.So, by parametrizing the line through the origin with direction vector (vec{e}), each coordinate is a linear function of t, and since t is the same for all coordinates, solving for t in each gives the symmetric equations where each coordinate is divided by its corresponding direction cosine. That makes sense.Let me think if there's another way to approach this. Maybe using projections. The direction cosines are the projections of the unit vector onto each axis, so any point on the line must have coordinates that are scalar multiples of these projections. Therefore, the ratios of the coordinates to the direction cosines must be equal, which is exactly the symmetric form of the line equation.Another check: suppose the direction vector was not a unit vector. For example, if the direction vector was (vec{v} = (a, b, c)), then the parametric equations would be (x = at), (y = bt), (z = ct), leading to the symmetric equations (frac{x}{a} = frac{y}{b} = frac{z}{c}). In our case, since it's a unit vector, (a = cos alpha), (b = cos beta), (c = cos gamma), so the symmetric equations are as given. Therefore, the logic holds.Therefore, converting the vector equation (vec{r} = tvec{e}) into coordinate form does result in (frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}). I don't see any mistakes here.Wait a second, in the problem statement, they wrote (z^{1}). That's probably a typo, right? Because z to the first power is just z. Maybe a formatting error when copying. The user probably meant (frac{z}{cos gamma}), which is standard. So that's fine.Is there a case where the direction cosines are zero? For example, if the line is along one of the coordinate axes. Suppose (alpha = 0), so the line is along the x-axis. Then (cos alpha = 1), and (cos beta = cos 90^circ = 0), same with (cos gamma = 0). Then the symmetric equations would be (frac{x}{1} = frac{y}{0} = frac{z}{0}), which is undefined for y and z. But in reality, the line along the x-axis is given by y = 0, z = 0. But in symmetric form, when direction cosines are zero, those denominators are zero, which isn't allowed. So, perhaps the symmetric form is only valid when all direction cosines are non-zero. But in cases where direction cosines are zero, the equations would be written separately. For example, if the line is along the x-axis, the equations would be y = 0 and z = 0, with x being free. However, in our case, the problem states that the direction vector forms angles with all three coordinate axes. If the line is along a coordinate axis, then two angles would be 90 degrees, and their cosines would be zero. But since (vec{e}) is a unit vector, the sum of squares of direction cosines must be 1. If two are zero, then the remaining one must be 1 or -1. So, in such cases, the symmetric form breaks down because division by zero occurs. So, perhaps the given coordinate form is only valid when none of the direction cosines are zero, i.e., the line is not parallel to any coordinate axis. However, the problem statement doesn't specify that, so maybe it's assumed that all the angles are such that their cosines are non-zero. Alternatively, in the original problem, if the line passes through the origin and is determined by a unit vector with direction angles (alpha), (beta), (gamma), then unless specified otherwise, we can assume that none of the cosines are zero. So, in the general case, the symmetric equations hold.Therefore, I think the reasoning is correct, and the conversion from the vector equation to the coordinate form is as shown. The key steps are recognizing that the direction vector components are the direction cosines, writing the parametric equations, then eliminating the parameter to get the symmetric form. It all checks out.Final AnswerThe coordinate form of the line's equation is boxed{frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}}.</think>The line ( l ) passes through the origin and its direction is determined by the unit vector (vec{e}), which forms angles (alpha), (beta), and (gamma) with the coordinate axes (Ox), (Oy), and (Oz) respectively. The vector equation of the line is given by (vec{r} = tvec{e}), where (t) is a parameter.The unit vector (vec{e}) can be written in terms of its direction cosines as (vec{e} = (cos alpha, cos beta, cos gamma)). Substituting this into the vector equation, we get the parametric equations for the coordinates of any point (M) on the line:[x = t cos alpha y = t cos beta z = t cos gamma]To convert these parametric equations into the coordinate form, we solve each equation for the parameter (t):[t = frac{x}{cos alpha} t = frac{y}{cos beta} t = frac{z}{cos gamma}]Since all three expressions equal (t), they must be equal to each other. Therefore, we have:[frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}]This is the symmetric form of the line's equation in coordinate form. The final answer is:[boxed{frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}}],<think>Okay, let's see. I need to convert the vector equation of the line, which is given as (vec{r} = tvec{e}), into the coordinate form that's shown as (frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}). Hmm, so first, let me make sure I understand all the components here.The line (l) passes through the origin, right? So that's why the vector equation is (vec{r} = tvec{e}). Since it's passing through the origin, there's no constant term; it's just a scalar multiple of the direction vector. The direction is determined by the unit vector (vec{e}), which has direction cosines (cos alpha), (cos beta), and (cos gamma) with respect to the coordinate axes (Ox), (Oy), and (Oz). So, the unit vector (vec{e}) can be written in component form as ((cos alpha, cos beta, cos gamma)), right? Because the direction cosines are the components of the unit vector in 3D space. Let me verify that. If a vector makes angles (alpha), (beta), and (gamma) with the x, y, and z-axes respectively, then the components of the unit vector in those directions are indeed the cosines of those angles. Yeah, that sounds familiar from the direction cosines concept.Therefore, the vector equation (vec{r} = tvec{e}) would translate component-wise to coordinates. Since (vec{r}) is the position vector ((x, y, z)), and (vec{e}) is ((cos alpha, cos beta, cos gamma)), substituting into the equation gives:[(x, y, z) = t (cos alpha, cos beta, cos gamma)]Which means each coordinate is equal to (t) times the respective direction cosine. So:[x = t cos alpha y = t cos beta z = t cos gamma]Now, to convert this into the coordinate form, which is the symmetric equation, we need to eliminate the parameter (t). Since each coordinate is expressed in terms of (t), we can solve each equation for (t) and set them equal to each other. From the first equation: (t = frac{x}{cos alpha})From the second: (t = frac{y}{cos beta})From the third: (t = frac{z}{cos gamma})Since all three expressions equal (t), they must equal each other. Therefore:[frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}]That seems straightforward. Let me check if there are any conditions or restrictions. Since (vec{e}) is a unit vector, the direction cosines satisfy the relation:[cos^2 alpha + cos^2 beta + cos^2 gamma = 1]But I don't think that affects the equation of the line. The key is that each coordinate divided by the corresponding direction cosine equals the same parameter (t), so setting them equal gives the symmetric equations of the line. Wait a second, is there a case where a direction cosine could be zero? For example, if the line is along one of the coordinate axes, say the x-axis. Then (alpha = 0), so (cos alpha = 1), and (beta = gamma = 90^circ), so (cos beta = 0), (cos gamma = 0). Then the equations would be (x = t), (y = 0), (z = 0), which is the x-axis. But in the symmetric form, we would have denominators with zero for y and z. So in that case, the symmetric equations are written as (y/0 = z/0), which is undefined. Hmm, but maybe the original problem assumes that none of the direction cosines are zero? Or perhaps in the coordinate form, we just write the ratios even if some denominators are zero, understanding that the corresponding numerators must be zero as well.But in the given problem statement, they present the equation as (frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}), so probably they are assuming that (cos alpha), (cos beta), and (cos gamma) are all non-zero. That would make sense if the line isn't parallel to any coordinate plane. But in general, the symmetric equations can handle cases where some denominators are zero by considering the corresponding numerators must also be zero. However, in this specific problem, since the direction is given by a unit vector with direction cosines, maybe all the angles are between 0 and 180 degrees, and none of the direction cosines are zero. The problem doesn't specify, but since it wants the conversion to that form, maybe it's safe to proceed under the assumption that none of the cosines are zero.Therefore, the conclusion is that by expressing each coordinate in terms of the parameter (t) and then equating the expressions for (t), we derive the symmetric equations of the line. So the coordinate form is indeed (frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}).Just to recap:1. Start with vector equation (vec{r} = tvec{e}).2. Express (vec{e}) in terms of direction cosines: ((cos alpha, cos beta, cos gamma)).3. Write component-wise equations: (x = t cos alpha), (y = t cos beta), (z = t cos gamma).4. Solve each equation for (t): (t = x / cos alpha = y / cos beta = z / cos gamma).5. Therefore, the symmetric equations are the ratios given.I think that's all there is to it. No steps seem missing. The key idea is understanding that the direction vector is a unit vector expressed via direction cosines, and translating the parametric equations (from the vector form) into symmetric form by eliminating the parameter. Makes sense.Final AnswerThe coordinate form of the line is boxed{dfrac{x}{cos alpha} = dfrac{y}{cos beta} = dfrac{z}{cos gamma}}.</think>The line ( l ) passes through the origin and has its direction determined by the unit vector (vec{e}), which forms angles (alpha), (beta), and (gamma) with the coordinate axes (Ox), (Oy), and (Oz). The vector equation of the line is given by (vec{r} = tvec{e}), where (vec{r} = overrightarrow{OM}) is the radius vector of any point (M) on the line, and (t) is a parameter.The unit vector (vec{e}) can be expressed in terms of direction cosines as ((cos alpha, cos beta, cos gamma)). Substituting this into the vector equation, we get the component-wise equations:[x = t cos alpha, quad y = t cos beta, quad z = t cos gamma.]To convert these parametric equations into the coordinate form, we solve each equation for (t):[t = frac{x}{cos alpha}, quad t = frac{y}{cos beta}, quad t = frac{z}{cos gamma}.]Since all three expressions are equal to (t), we set them equal to each other, resulting in the symmetric equations:[frac{x}{cos alpha} = frac{y}{cos beta} = frac{z}{cos gamma}.]Thus, the coordinate form of the line is boxed{dfrac{x}{cos alpha} = dfrac{y}{cos beta} = dfrac{z}{cos gamma}}.]

🔑:40. Let cos alpha, cos beta, cos gamma be the direction cosines of the unit vector vec{e}. Since vectors vec{r} and vec{e} are collinear, from the equality x vec{i} + 132+y vec{j}+z vec{k}=t vec{i} cos alpha+t vec{j} cos beta+t vec{k} cos gamma quad we obtain: quad x=t cos alpha, y=t cos beta, z=t cos gamma ; frac{x}{cos alpha}=frac{y}{cos beta}=frac{z}{cos gamma}.